- A zero velocity vector requires both components to be zero simultaneously. - Examine whether either component can ever vanish. - One component may settle the question without solving the other equation.

1. A stopped particle must have both velocity components equal to \(0\). 2. The horizontal component \(t^2+1\) is positive for every real \(t\). 3. Therefore, the velocity vector can never be \(\langle0,0\rangle\), so the particle never stops.

The particle never stops.

- Velocity describes the instantaneous rate of change of position. - Treat the horizontal and vertical position components separately. - Evaluate the resulting rate vector at the stated time.

1. Differentiate the position components: \(\mathbf v(t)=\mathbf r'(t)=\langle2t,3\rangle\). 2. Evaluate at \(t=1\): \(\mathbf v(1)=\langle2,3\rangle\).

\(\mathbf v(1)=\langle2,3\rangle\)

- Velocity is a vector, while speed is a nonnegative scalar. - Think about how the magnitude of a two-component vector is found. - The sign of a velocity component affects direction but not its squared contribution to speed.

1. Speed is the magnitude of the velocity vector. 2. Thus, \(\|\mathbf v\|=\sqrt{3^2+(-4)^2}=5\).

\(5\)

- Speed combines horizontal and vertical rates, not their simple sum. - Use the magnitude of the two-component rate vector.

1. The two coordinate rates are perpendicular components of the rate of motion. 2. Their magnitude is \(\sqrt{3^2+4^2}=5\).

\(5\) coordinate units per unit of \(t\).

- Distance traveled follows both pieces of the drawn route. - The endpoint distance is the length of the direct segment from P to R.

1. The skater travels \(3\) units horizontally and \(4\) units vertically, a total of \(7\). 2. The straight-line displacement magnitude is \(\sqrt{3^2+4^2}=5\). 3. Total distance traveled is greater.

Total distance traveled (\(7\) units versus \(5\) units).

- A velocity component may be negative even though speed cannot be. - Speed is the magnitude of the vector.

1. The horizontal velocity component is negative, so the cart moves left. 2. The magnitude of the vector is \(\sqrt{(-5)^2+0^2}=5\) meters per second.

Speed \(5\,\text{m/s}\), moving left.

- Acceleration describes how velocity changes. - Differentiate both vector components, even the zero one.

1. Acceleration is the derivative of velocity. 2. Differentiate each component: \(\mathbf{a}(t)=\langle1,0\rangle\) meters per second squared.

\(\mathbf{a}(t)=\langle1,0\rangle\,\text{m/s}^2\).

- Differentiate the position vector once for velocity and twice for acceleration. - Substitute the requested time only after finding the derivative functions. - Compute speed as the nonnegative magnitude of the velocity vector.

1. Differentiate position to get \(\mathbf{v}(t)=\mathbf{r}'(t)=\left\langle 2t, 3\right\rangle\), so \(\mathbf{v}(2)=\left\langle 4, 3\right\rangle\). 2. Differentiate velocity to get \(\mathbf{a}(t)=\mathbf{v}'(t)=\left\langle 2, 0\right\rangle\), so \(\mathbf{a}(2)=\left\langle 2, 0\right\rangle\). 3. The speed is \(\|\mathbf{v}(2)\|=\sqrt{4^{2}+3^{2}}=5\).

Velocity: \(\left\langle 4, 3\right\rangle\); acceleration: \(\left\langle 2, 0\right\rangle\); speed: \(5\)

- Differentiate the coordinate functions once for velocity and again for acceleration. - Use the exact sine and cosine values at \(\frac{\pi}{3}\). - Find speed from the magnitude of the evaluated velocity, not from either component alone.

1. The velocity is \(\mathbf{v}(t)=\left\langle -\sin(t), \cos(t)\right\rangle\), so \(\mathbf{v}\left(\frac{\pi}{3}\right)=\left\langle -\frac{\sqrt{3}}{2}, \frac{1}{2}\right\rangle\). 2. The acceleration is \(\mathbf{a}(t)=\left\langle -\cos(t), -\sin(t)\right\rangle\), so \(\mathbf{a}\left(\frac{\pi}{3}\right)=\left\langle -\frac{1}{2}, -\frac{\sqrt{3}}{2}\right\rangle\). 3. The speed is \(\left\|\mathbf{v}\left(\frac{\pi}{3}\right)\right\|=\sqrt{\frac{3}{4}+\frac{1}{4}}=1\).

Velocity: \(\left\langle -\frac{\sqrt{3}}{2}, \frac{1}{2}\right\rangle\); acceleration: \(\left\langle -\frac{1}{2}, -\frac{\sqrt{3}}{2}\right\rangle\); speed: \(1\)

- Differentiate the position vector once for velocity and twice for acceleration. - Evaluate the derivative vectors at the requested time. - Compute speed from the magnitude of velocity, not acceleration.

1. The velocity is \(\mathbf{v}(t)=\mathbf{r}'(t)=\left\langle e^{t}, 2t\right\rangle\), so \(\mathbf{v}(0)=\left\langle 1, 0\right\rangle\). 2. The acceleration is \(\mathbf{a}(t)=\mathbf{v}'(t)=\left\langle e^{t}, 2\right\rangle\), so \(\mathbf{a}(0)=\left\langle 1, 2\right\rangle\). 3. The speed is \(\|\mathbf{v}(0)\|=\sqrt{1^{2}+0^{2}}=1\).

Velocity: \(\left\langle 1, 0\right\rangle\); acceleration: \(\left\langle 1, 2\right\rangle\); speed: \(1\)

- Simplify or apply the product rule to the first position component before differentiating. - Differentiate once for velocity and a second time for acceleration. - Use the evaluated velocity components to calculate the nonnegative speed.

1. The velocity is \(\mathbf{v}(t)=\left\langle 3t^{2}-3, 2t\right\rangle\), so \(\mathbf{v}(1)=\left\langle 0, 2\right\rangle\). 2. The acceleration is \(\mathbf{a}(t)=\left\langle 6t, 2\right\rangle\), so \(\mathbf{a}(1)=\left\langle 6, 2\right\rangle\). 3. The speed is \(\|\mathbf{v}(1)\|=\sqrt{0^{2}+2^{2}}=2\).

Velocity: \(\left\langle 0, 2\right\rangle\); acceleration: \(\left\langle 6, 2\right\rangle\); speed: \(2\)

- Rewrite reciprocal powers with negative exponents before differentiating if helpful. - Differentiate each component twice, keeping the coordinate order fixed. - Find speed by taking the magnitude of the velocity at \(t=1\).

1. The velocity is \(\mathbf{v}(t)=\left\langle \frac{1}{t}, -\frac{1}{t^{2}}\right\rangle\), so \(\mathbf{v}(1)=\left\langle 1, -1\right\rangle\). 2. The acceleration is \(\mathbf{a}(t)=\left\langle -\frac{1}{t^{2}}, \frac{2}{t^{3}}\right\rangle\), so \(\mathbf{a}(1)=\left\langle -1, 2\right\rangle\). 3. The speed is \(\|\mathbf{v}(1)\|=\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}\).

Velocity: \(\left\langle 1, -1\right\rangle\); acceleration: \(\left\langle -1, 2\right\rangle\); speed: \(\sqrt{2}\)

- Differentiate each trigonometric coordinate carefully, including its sign. - Use the exact sine and cosine values at \(t=\pi\). - Calculate speed from the evaluated velocity vector only.

1. The velocity is \(\mathbf{v}(t)=\left\langle 1-\cos(t), \sin(t)\right\rangle\), so \(\mathbf{v}(\pi)=\left\langle 2, 0\right\rangle\). 2. The acceleration is \(\mathbf{a}(t)=\left\langle \sin(t), \cos(t)\right\rangle\), so \(\mathbf{a}(\pi)=\left\langle 0, -1\right\rangle\). 3. The speed is \(\|\mathbf{v}(\pi)\|=\sqrt{2^{2}+0^{2}}=2\).

Velocity: \(\left\langle 2, 0\right\rangle\); acceleration: \(\left\langle 0, -1\right\rangle\); speed: \(2\)

- Displacement is the definite integral of velocity over the time interval. - Integrate the velocity components separately using the same bounds. - Add the displacement vector to the initial position to obtain the final position.

1. The displacement from \(t=0\) to \(t=4\) is \(\int_{0}^{4}\mathbf{v}(t)\,dt=\left\langle 16, 12\right\rangle\). 2. Adding the displacement to the initial position gives \(\mathbf{r}(4)=\left\langle 1, -2\right\rangle+\left\langle 16, 12\right\rangle=\left\langle 17, 10\right\rangle\).

Displacement: \(\left\langle 16, 12\right\rangle\); position at \(t=4\): \(\left\langle 17, 10\right\rangle\)

- Integrate the velocity vector over the interval to find displacement. - Evaluate the sine and cosine antiderivatives at both endpoints. - Add the displacement vector to the given initial position.

1. The displacement is \(\int_{0}^{\pi/2}\mathbf{v}(t)\,dt=\left\langle 1, 1\right\rangle\). 2. Adding displacement to the initial position gives \(\mathbf{r}\left(\frac{\pi}{2}\right)=\left\langle 2, 1\right\rangle+\left\langle 1, 1\right\rangle=\left\langle 3, 2\right\rangle\).

Displacement: \(\left\langle 1, 1\right\rangle\); position at \(t=\frac{\pi}{2}\): \(\left\langle 3, 2\right\rangle\)

- Use the definite integral of velocity from the initial time to the final time. - Integrate the polynomial components separately with bounds \(1\) and \(3\). - Add the resulting displacement to \(\mathbf{r}(1)\) component by component.

1. The displacement is \(\int_{1}^{3}\mathbf{v}(t)\,dt=\left\langle \frac{26}{3}, 8\right\rangle\). 2. Adding displacement to the initial position gives \(\mathbf{r}(3)=\left\langle 0, 5\right\rangle+\left\langle \frac{26}{3}, 8\right\rangle=\left\langle \frac{26}{3}, 13\right\rangle\).

Displacement: \(\left\langle \frac{26}{3}, 8\right\rangle\); position at \(t=3\): \(\left\langle \frac{26}{3}, 13\right\rangle\)

- Integrate both velocity components over \([0,1]\) to obtain displacement. - Preserve the negative sign in the constant vertical velocity. - Add the displacement to the initial vector to find the later position.

1. The displacement is \(\int_{0}^{1}\mathbf{v}(t)\,dt=\left\langle e-1, -1\right\rangle\). 2. Adding displacement to the initial position gives \(\mathbf{r}(1)=\left\langle 1, 4\right\rangle+\left\langle e-1, -1\right\rangle=\left\langle e, 3\right\rangle\).

Displacement: \(\left\langle e-1, -1\right\rangle\); position at \(t=1\): \(\left\langle e, 3\right\rangle\)

- Set the displacement bounds from the given initial time to the requested final time. - Evaluate each polynomial antiderivative carefully at the negative lower bound. - Add the displacement vector to the known position at \(t=-1\).

1. The displacement is \(\int_{-1}^{2}\mathbf{v}(t)\,dt=\left\langle 3, 6\right\rangle\). 2. Adding displacement to the initial position gives \(\mathbf{r}(2)=\left\langle 5, 0\right\rangle+\left\langle 3, 6\right\rangle=\left\langle 8, 6\right\rangle\).

Displacement: \(\left\langle 3, 6\right\rangle\); position at \(t=2\): \(\left\langle 8, 6\right\rangle\)

- Total distance is found by integrating the magnitude of velocity. - Simplify the speed using a fundamental trigonometric identity. - Integrate the resulting constant over the full time interval.

1. The speed is \(\|\mathbf{v}(t)\|=\sqrt{\cos^{2}(t)+\sin^{2}(t)}=1\). 2. Therefore, the total distance is \(\int_{0}^{\pi}1\,dt=\pi\).

\(\pi\)

- A stopped particle must have a zero velocity vector. - Set both velocity components equal to zero independently. - Check that the same parameter value solves both component equations.

1. The particle is stopped when \(\mathbf{v}(t)=\left\langle 0, 0\right\rangle\), so both component equations must hold. 2. Solving \(2t-4=0\) and \(3t-6=0\) gives the common time \(t=2\).

\(t=2\)

- Differentiate the position components to obtain velocity. - Use the signs of the evaluated velocity components to determine horizontal and vertical direction. - Compute speed as the magnitude of the velocity vector.

1. The velocity is \(\mathbf{v}(t)=\left\langle 2t, 3t^{2}-4\right\rangle\). 2. At \(t=1\), \(\mathbf{v}(1)=\left\langle 2, -1\right\rangle\). 3. The positive first component and negative second component mean the particle is moving right and down. 4. Its speed is \(\|\mathbf{v}(1)\|=\sqrt{2^{2}+(-1)^{2}}=\sqrt{5}\).

Moving right and down; speed: \(\sqrt{5}\)

- Compare the endpoint positions before computing distance. - Differentiate the position to find speed. - Constant speed simplifies the distance integral.

1. The initial and final positions are both \((3,0)\), so the displacement is \(\langle0,0\rangle\). 2. The velocity is \(\mathbf{v}(t)=\langle-6\sin(2t),6\cos(2t)\rangle\), so the speed is \(6\). 3. The total distance is \(\int_0^{2\pi}6\,dt=12\pi\).

Displacement: \(\langle0,0\rangle\); total distance: \(12\pi\)

- Directly rightward motion imposes both a zero-component and a sign condition. - Once the time is found, evaluate the velocity magnitude. - Differentiate velocity for the acceleration vector.

1. Directly rightward motion requires the vertical velocity to be \(0\) and the horizontal velocity to be positive. 2. The vertical component is \(t-1\), so \(t=1\). Then the horizontal component is \(2>0\). 3. The speed at that time is \(\|\langle2,0\rangle\|=2\). 4. The acceleration is \(\mathbf{a}(t)=\langle2t,1\rangle\), so \(\mathbf{a}(1)=\langle2,1\rangle\).

At \(t=1\), the particle moves directly right with speed \(2\) and acceleration \(\langle2,1\rangle\).

- Estimate the derivative of velocity using data equally spaced around the target time. - Subtract the earlier velocity vector from the later one. - Divide the entire change vector by the elapsed time.

1. A symmetric estimate is \(\mathbf{a}(2)\approx\frac{\mathbf{v}(2.1)-\mathbf{v}(1.9)}{2.1-1.9}\). 2. The velocity change is \(\langle0.80,0.20\rangle\), and the time change is \(0.20\). 3. Dividing componentwise gives \(\mathbf{a}(2)\approx\langle4,1\rangle\).

\(\mathbf{a}(2)\approx\langle4,1\rangle\)

- Compute displacement separately on each time interval. - Direction affects displacement but not the nonnegative distance contribution. - Add vectors for displacement and scalar lengths for distance.

1. From \(t=0\) to \(t=2\), the displacement is \(2\langle1,0\rangle=\langle2,0\rangle\), and the distance is \(2\). 2. From \(t=2\) to \(t=3\), the displacement is \(1\langle-2,0\rangle=\langle-2,0\rangle\), and the distance is \(2\). 3. The total displacement is \(\langle0,0\rangle\), while the total distance is \(4\).

Displacement: \(\langle0,0\rangle\); total distance: \(4\)

- Use the horizontal position coordinate to locate a y-axis crossing. - Apply the time-domain restriction. - Evaluate the signs of both velocity components at the crossing time.

1. The particle is on the y-axis when \(t^2-4=0\). On \(t\ge0\), this gives \(t=2\). 2. The crossing point is \(\mathbf{r}(2)=(0,2)\). 3. The velocity is \(\mathbf{v}(t)=\langle2t,3t^2-3\rangle\), so \(\mathbf{v}(2)=\langle4,9\rangle\). 4. Both components are positive, so the particle is moving right and upward.

At \(t=2\), the particle crosses the y-axis at \((0,2)\), moving right and upward.

- A position graph's slope gives the sign of that coordinate's velocity. - Check the direction of each graph as time increases through \(t=1\). - Decide horizontal and vertical motion separately.

1. The position graph \(x(t)\) rises at \(t=1\), so \(\frac{dx}{dt}>0\) and the particle moves right. 2. The position graph \(y(t)\) falls at \(t=1\), so \(\frac{dy}{dt}<0\) and the particle moves down.

Right and down.

- Read each marked time from the horizontal axis. - Use the two drawn straight lines to estimate the instantaneous slopes. - Check the estimates using the derivative of the given position function.

1. From the graph, P occurs at \(t=1\) and Q at \(t=2\). 2. The straight line through P has slope \(\frac{2-0}{1.5-0.5}=2\,\text{m/s}\). The line through Q has slope \(\frac{5-3}{2.25-1.75}=4\,\text{m/s}\). 3. Differentiating gives \(\frac{dx}{dt}=2t\), so the exact velocities are \(2\,\text{m/s}\) at P and \(4\,\text{m/s}\) at Q. 4. Q has the greater velocity, agreeing with the steeper tangent line.

P: \(t=1\), velocity \(2\,\text{m/s}\). Q: \(t=2\), velocity \(4\,\text{m/s}\). Q is faster; both graph estimates match the derivative.

- Read each velocity component at the same time. - Combine perpendicular components by finding their vector magnitude.

1. At \(t=1\), the graphs give \(v_x=2\) and \(v_y=1\). 2. Speed is the magnitude: \(\sqrt{2^2+1^2}=\sqrt{5}\) meters per second.

\(\sqrt{5}\,\text{m/s}\).

- Each velocity component changes its coordinate for the full two seconds. - Add displacement to the initial position rather than reporting displacement alone.

1. Multiply velocity by time to obtain displacement \(\langle-2,6\rangle\) meters. 2. Add to the starting coordinates: \((1-2,2+6)=(-1,8)\).

\((-1,8)\) meters.

- Distinguish the direction of travel from the direction of acceleration. - What happens when acceleration points against the current velocity?

1. The train moves in the positive horizontal direction but accelerates in the opposite direction. 2. Its speed is decreasing at that instant.

Decreasing.

- First convert the total displacement from \(A\) to \(B\) into displacement per second. - The position function starts at \(A\) and adds the elapsed-time multiple of the constant velocity. - Use the altitude component to determine the requested time. - For part c), distinguish the velocity vector from its scalar magnitude.

1. The displacement from \(A\) to \(B\) is \(\langle30,30,-30\rangle\). Over \(6\,\text{s}\), the constant velocity is \(\langle5,5,-5\rangle\,\text{m/s}\). 2. Therefore, \(\mathbf{r}(t)=\langle10+5t,20+5t,100-5t\rangle\) for \(0\le t\le6\). 3. Set the altitude equal to \(85\): \(100-5t=85\), so \(t=3\,\text{s}\). Then \(\mathbf{r}(3)=\langle25,35,85\rangle\), giving position \((25, 35, 85)\). 4. Differentiating componentwise gives \(\mathbf{r}'(t)=\langle5,5,-5\rangle\,\text{m/s}\). Its magnitude is \(\sqrt{5^2+5^2+(-5)^2}=5\sqrt{3}\,\text{m/s}\).

a) \(\mathbf{r}(t)=\langle10+5t,20+5t,100-5t\rangle\), \(0\le t\le6\) b) \(t=3\,\text{s}\), at \((25, 35, 85)\) c) Velocity \(\langle5,5,-5\rangle\,\text{m/s}\); speed \(5\sqrt{3}\,\text{m/s}\)

- Total distance is the integral of the velocity vector's magnitude. - Simplify the square root using the relationship between the two velocity components. - Use the interval restriction to resolve the absolute value before integrating.

1. The speed is \(\|\mathbf{v}(t)\|=\sqrt{(3t)^{2}+(4t)^{2}}=5|t|\). 2. Since \(t\ge0\) on the interval, the speed is \(5t\). 3. The total distance is \(\int_{0}^{5}5t\,dt=\frac{125}{2}\).

\(\frac{125}{2}\)

- Form the speed by taking the magnitude of the velocity vector. - Use the nonnegative time interval to simplify the absolute-value factor. - Look for a substitution based on the expression inside the square root.

1. The speed is \(\|\mathbf{v}(t)\|=\sqrt{(2t)^{2}+(t^{2})^{2}}=|t|\sqrt{t^{2}+4}\). 2. Since \(t\ge0\), the speed is \(t\sqrt{t^{2}+4}\). 3. Using \(u=t^{2}+4\), the distance is \(\int_{0}^{2}t\sqrt{t^{2}+4}\,dt=\frac{1}{3}\left[(t^{2}+4)^{3/2}\right]_{0}^{2}=\frac{16\sqrt{2}-8}{3}\).

\(\frac{16\sqrt{2}-8}{3}\)

- Build the speed from both velocity components before integrating. - The resulting definite integral does not need to be simplified to an elementary antiderivative. - Keep full calculator precision until rounding the final distance.

1. The speed is \(\|\mathbf{v}(t)\|=\sqrt{1+\cos^{2}(t)}\). 2. The total distance is \(\int_{0}^{\pi}\sqrt{1+\cos^{2}(t)}\,dt\). 3. Numerical evaluation gives \(3.820197789\ldots\), so the distance is \(\approx3.820\).

\(\approx3.820\)

- Square both velocity components when forming the speed. - Set up the definite integral over the entire time interval before using numerical evaluation. - Round only the final distance, not intermediate values.

1. The speed is \(\|\mathbf{v}(t)\|=\sqrt{e^{2t}+t^{2}}\). 2. The total distance is \(\int_{0}^{1}\sqrt{e^{2t}+t^{2}}\,dt\). 3. Numerical evaluation gives \(1.796429783\ldots\), so the distance is \(\approx1.796\).

\(\approx1.796\)

- Minimize the squared speed to avoid differentiating a square root. - Express the squared magnitude as a quadratic in \(t\). - Check that the quadratic's minimizing time lies in the stated domain before taking the square root.

1. Since the square-root function is increasing, minimize the squared speed: \(\|\mathbf{v}(t)\|^{2}=t^{2}+(2-t)^{2}=2t^{2}-4t+4\). 2. Completing the square gives \(2(t-1)^{2}+2\), whose minimum on \(t\ge0\) occurs at \(t=1\). 3. The minimum speed is \(\sqrt{2}\).

\(t=1\); minimum speed: \(\sqrt{2}\)

- Work with the square of the speed, which has the same minimizing time as the speed. - Expand or complete the square for the resulting quadratic. - After finding the minimum squared speed, take its nonnegative square root.

1. Minimize the squared speed: \(\|\mathbf{v}(t)\|^{2}=(2t-3)^{2}+(t+1)^{2}=5t^{2}-10t+10\). 2. Completing the square gives \(5(t-1)^{2}+5\), whose minimum on \(t\ge0\) occurs at \(t=1\). 3. The minimum speed is \(\sqrt{5}\).

\(t=1\); minimum speed: \(\sqrt{5}\)

- Identify the type of quantity displacement should be and the type of quantity distance should be. - Ask what the value \(14\) is actually doing to horizontal and vertical information. - Compute each requested quantity from its meaning, then compare with the claim.

1. The displacement is the vector integral \(\int_0^2\mathbf{v}(t)\,\text{d}t=\left\langle6,8\right\rangle\). 2. The speed is the scalar magnitude \(\|\mathbf{v}(t)\|=\sqrt{3^2+4^2}=5\). 3. The distance is \(\int_0^2 5\,\text{d}t=10\). Adding vector components does not produce either displacement or distance.

Displacement: \(\left\langle6,8\right\rangle\); distance: \(10\). The value \(14\) incorrectly adds vector components.

- The proposed procedure combines net coordinate changes; decide whether that matches Euclidean path length. - Separate the vector question “Where did the particle end relative to where it began?” from the scalar question “How far did it travel?” - Use the geometry of a two-component velocity vector when constructing the scalar travel rate.

1. The displacement is \(\int_0^2\mathbf{v}(t)\,\text{d}t=\left\langle2,-2\right\rangle\). 2. Adding the absolute values of the net component changes gives a taxicab-style length, not the particle's path length. 3. The speed is \(\|\mathbf{v}(t)\|=\sqrt{t^2+(-t)^2}=\sqrt2\,t\) on \([0,2]\). 4. Therefore, the distance is \(\int_0^2\sqrt2\,t\,\text{d}t=2\sqrt2\).

Displacement: \(\left\langle2,-2\right\rangle\); distance: \(2\sqrt2\). Adding absolute component changes does not compute Euclidean path length.

- Inspect the claimed “speed”: is it a scalar or a vector? - Distinguish the instantaneous magnitude of velocity from the accumulated velocity vector. - Once the correct speed is identified, use it to determine total distance over the interval.

1. The vector \(\int_0^{\pi/2}\mathbf{v}(t)\,\text{d}t=\left\langle1,1\right\rangle\) is the displacement, not the speed. 2. The speed is the scalar \(\|\mathbf{v}(t)\|=\sqrt{\cos^2t+\sin^2t}=1\). 3. The distance is \(\int_0^{\pi/2}1\,\text{d}t=\frac{\pi}{2}\).

Displacement: \(\left\langle1,1\right\rangle\); speed: \(1\); distance: \(\frac{\pi}{2}\)

- Minimize the squared distance from the position vector to the origin. - Rewrite the resulting quadratic in vertex form or locate its critical point. - Verify the minimizing time is allowed, then take the square root of the minimum squared distance.

1. Minimize the squared distance \(D^{2}(t)=(t-2)^{2}+(t+1)^{2}=2t^{2}-2t+5\). 2. Completing the square gives \(D^{2}(t)=2\left(t-\frac{1}{2}\right)^{2}+\frac{9}{2}\), so the minimum on \(t\ge0\) occurs at \(t=\frac{1}{2}\). 3. The minimum distance is \(\sqrt{\frac{9}{2}}=\frac{3\sqrt{2}}{2}\).

\(t=\frac{1}{2}\); minimum distance: \(\frac{3\sqrt{2}}{2}\)

- Approximate horizontal and vertical displacement from their velocity columns separately. - Use the spacing in the time column as the trapezoidal-rule width. - Add the approximate displacement vector to the given starting position.

1. The subinterval width is \(1\). For the horizontal displacement, the trapezoidal rule gives \(\frac{1}{2}[(1+2)+(2+0)+(0-1)]=2\). 2. For the vertical displacement, it gives \(\frac{1}{2}[(0+2)+(2+3)+(3+1)]=\frac{11}{2}\). 3. Thus, the displacement is approximately \(\left\langle 2, \frac{11}{2}\right\rangle\). 4. Adding the initial position gives \(\mathbf{r}(3)\approx\left\langle 4, -2\right\rangle+\left\langle 2, \frac{11}{2}\right\rangle=\left\langle 6, \frac{7}{2}\right\rangle\).

\(\mathbf{r}(3)\approx\left\langle 6, \frac{7}{2}\right\rangle\)

- Use one trapezoidal approximation for each velocity component. - Determine the common subinterval width from consecutive time values. - Interpret the two approximations as displacement and add them to the initial coordinates.

1. The subinterval width is \(0.5\). For the horizontal displacement, the trapezoidal rule gives \(\frac{0.5}{2}[(0+2)+(2+4)+(4+2)]=\frac{7}{2}\). 2. For the vertical displacement, it gives \(\frac{0.5}{2}[(2+4)+(4+2)+(2+0)]=\frac{7}{2}\). 3. Thus, the displacement is approximately \(\left\langle \frac{7}{2}, \frac{7}{2}\right\rangle\). 4. Adding the initial position gives \(\mathbf{r}(1.5)\approx\left\langle 1, 1\right\rangle+\left\langle \frac{7}{2}, \frac{7}{2}\right\rangle=\left\langle \frac{9}{2}, \frac{9}{2}\right\rangle\).

\(\mathbf{r}(1.5)\approx\left\langle \frac{9}{2}, \frac{9}{2}\right\rangle\)

- Apply the trapezoidal rule independently to the horizontal and vertical velocity columns. - Interpret the two numerical integrals as components of displacement. - Add the displacement vector to the position at the initial time.

1. The subinterval width is \(1\). The horizontal displacement is approximately \(\frac{1}{2}[(3+1)+(1+2)+(2+0)]=\frac{9}{2}\). 2. The vertical displacement is approximately \(\frac{1}{2}[(-1+1)+(1+4)+(4+2)]=\frac{11}{2}\). 3. Thus, the estimated displacement is \(\Delta\mathbf{r}\approx\left\langle\frac{9}{2},\frac{11}{2}\right\rangle\). 4. Adding the initial position gives \(\mathbf{r}(4)\approx\left\langle-2,5\right\rangle+\left\langle\frac{9}{2},\frac{11}{2}\right\rangle=\left\langle\frac{5}{2},\frac{21}{2}\right\rangle\).

\(\mathbf{r}(4)\approx\left\langle\frac{5}{2},\frac{21}{2}\right\rangle\)

- Horizontal and vertical motion impose different zero-component conditions. - Apply the domain restriction after solving each condition. - Use the sign of the remaining component to determine direction.

1. Horizontal motion requires the vertical component to be \(0\): \(t^2-1=0\). On \(t\ge0\), this gives \(t=1\). 2. At \(t=1\), \(\mathbf{v}(1)=\langle-1,0\rangle\), so the particle moves left. 3. Vertical motion requires the horizontal component to be \(0\): \(t-2=0\), so \(t=2\). 4. At \(t=2\), \(\mathbf{v}(2)=\langle0,3\rangle\), so the particle moves upward.

At \(t=1\), the particle moves horizontally to the left. At \(t=2\), it moves vertically upward.

- A stopped particle has a zero velocity vector, not merely one zero component. - Find the common solution of the two component equations. - Differentiate the velocity vector to obtain acceleration.

1. The particle is stopped when both velocity components are \(0\). 2. The equations \(t^2-4=0\) and \(2t-4=0\) have the common nonnegative solution \(t=2\). 3. The acceleration is \(\mathbf{a}(t)=\mathbf{v}'(t)=\langle2t,2\rangle\). 4. Therefore, \(\mathbf{a}(2)=\langle4,2\rangle\).

The particle is stopped at \(t=2\), and \(\mathbf{a}(2)=\langle4,2\rangle\).

- Relate the rate of change of distance to the position and velocity vectors. - The sign of the relevant dot product determines whether the particle is moving closer or farther away. - Keep the magnitude of the position vector in the denominator.

1. At \(t=1\), the position is \(\langle-3,2\rangle\) and the velocity is \(\langle2,2\rangle\). 2. If \(D=\|\mathbf{r}\|\), then \(D'=\frac{\mathbf{r}\cdot\mathbf{v}}{\|\mathbf{r}\|}\). 3. The dot product is \((-3)(2)+(2)(2)=-2\), and \(\|\mathbf{r}(1)\|=\sqrt{13}\). 4. Thus \(D'(1)=-\frac{2}{\sqrt{13}}\), so the distance is decreasing.

The distance is decreasing at the rate \(\frac{2}{\sqrt{13}}\) units per unit time; equivalently, \(D'(1)=-\frac{2}{\sqrt{13}}\).

- Integrate acceleration to recover velocity first. - Use the initial velocity before integrating a second time. - Apply the initial position after finding the position antiderivative.

1. Integrating acceleration and using \(\mathbf{v}(0)\) gives \(\mathbf{v}(t)=\langle3t^2+1,3-2t\rangle\). 2. Integrating velocity and using \(\mathbf{r}(0)\) gives \(\mathbf{r}(t)=\langle t^3+t-2,3t-t^2+4\rangle\). 3. Evaluating at \(t=2\) gives \(\mathbf{r}(2)=\langle8,6\rangle\).

\(\mathbf{r}(2)=\langle8,6\rangle\)

- Compare squared speeds so that no radical is needed. - Use the speed at the initial time as the threshold. - Intersect the solution of the inequality with the stated time interval.

1. The initial speed is \(\|\mathbf{v}(0)\|=3\). 2. The squared speed is \((t^2-3)^2+4t^2=t^4-2t^2+9\). 3. The condition \(\|\mathbf{v}(t)\|<3\) is equivalent to \(t^4-2t^2<0\), or \(t^2(t^2-2)<0\). 4. On the stated interval, this holds for \(0<t<\sqrt{2}\), whose duration is \(\sqrt{2}\).

The speed is less than its initial speed for \(0<t<\sqrt{2}\). The total duration is \(\sqrt{2}\).

- The velocity points in the two panels supply separate component data. - Apply Simpson's rule to each component using the same subinterval width. - The integrals give displacement, which must be added to the initial position.

1. With four equal subintervals, \(h=1\). Apply Simpson's rule separately to both velocity components. 2. The estimated horizontal displacement is \(\frac{1}{3}[0+4(2)+2(4)+4(2)+0]=8\). 3. The estimated vertical displacement is \(\frac{1}{3}[1+4(0)+2(-1)+4(0)+1]=0\). 4. Add the estimated displacement to the initial position: \(\mathbf{r}(4)\approx\langle2,-1\rangle+\langle8,0\rangle=\langle10,-1\rangle\).

\(\mathbf{r}(4)\approx\langle10,-1\rangle\)

- Differentiate velocity to obtain acceleration. - Use a sign test involving velocity and acceleration rather than differentiating a radical. - Evaluate the sign at the specified time.

1. The acceleration is \(\mathbf{a}(t)=\langle1,2t\rangle\). 2. Speed is increasing when \(\mathbf{v}\cdot\mathbf{a}>0\) and decreasing when this dot product is negative. 3. At \(t=\frac{1}{2}\), the dot product is \(\frac{1}{2}+\left(-\frac{3}{4}\right)(1)=-\frac{1}{4}\). 4. Therefore, the speed is decreasing.

The speed is decreasing at \(t=\frac{1}{2}\).

- Differentiate velocity to obtain acceleration. - Use a perpendicular-vector condition. - Apply the positive-time restriction after solving the equation.

1. The acceleration is \(\mathbf{a}(t)=\langle1,2t\rangle\). 2. Perpendicularity requires \(\mathbf{v}(t)\cdot\mathbf{a}(t)=0\). 3. The dot product is \(t+2t(t^2-2)=t(2t^2-3)\). 4. Since \(t>0\), the solution is \(t=\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{2}\).

\(t=\frac{\sqrt{6}}{2}\)

- Use an improper vector integral to find the total displacement. - Add the displacement to the initial position. - Integrate the magnitude of velocity separately for distance.

1. The total displacement is \(\int_0^{\infty}\mathbf{v}(t)\,dt=\langle1,2\rangle\). 2. Therefore, the limiting position is \(\langle1,2\rangle\). 3. The speed is \(\sqrt{e^{-2t}+4e^{-2t}}=\sqrt{5}e^{-t}\). 4. The total distance is \(\int_0^{\infty}\sqrt{5}e^{-t}\,dt=\sqrt{5}\).

Limiting position: \(\langle1,2\rangle\); total distance: \(\sqrt{5}\)

- Horizontal direction depends only on the sign of the horizontal velocity component. - Use its zeros to divide the time domain into intervals. - Test the sign on each interval.

1. Horizontal direction changes can occur when \(v_x(t)=(t-1)(t-3)=0\), so the candidate times are \(t=1\) and \(t=3\). 2. For \(0\le t<1\), both factors are negative, so \(v_x>0\) and the particle moves right. 3. For \(1<t<3\), the factors have opposite signs, so \(v_x<0\) and the particle moves left. 4. For \(t>3\), both factors are positive, so \(v_x>0\) and the particle moves right.

The horizontal direction changes at \(t=1\) and \(t=3\). The particle moves right for \(0\le t<1\) and \(t>3\), and left for \(1<t<3\).

- Work with the square of the speed. - Rewrite it so that one nonnegative trigonometric term controls the minimum. - List every time in the closed interval that produces that minimum.

1. The squared speed is \(S(t)=4\cos^2t+\sin^2t=1+3\cos^2t\). 2. This expression is minimized when \(\cos t=0\). 3. On \([0,2\pi]\), this occurs at \(t=\frac{\pi}{2}\) and \(t=\frac{3\pi}{2}\). 4. The minimum speed is \(\sqrt{1}=1\).

The minimum speed is \(1\), occurring at \(t=\frac{\pi}{2}\) and \(t=\frac{3\pi}{2}\).

- Differentiate velocity to obtain acceleration. - Use a parallel-vector condition in the plane. - Compare the vectors at the valid time to determine orientation.

1. The acceleration is \(\mathbf{a}(t)=\langle1,2t\rangle\). 2. Parallel planar vectors have determinant \(0\): \(t(2t)-(t^2+1)(1)=t^2-1=0\). 3. On \(t\ge0\), the solution is \(t=1\). 4. At \(t=1\), both vectors equal \(\langle1,2\rangle\), so they point in the same direction.

At \(t=1\), the velocity and acceleration vectors are parallel and point in the same direction.

- Differentiate velocity to obtain acceleration. - Relate the derivative of speed to the velocity–acceleration dot product. - Evaluate both the dot product and speed at the stated time.

1. The acceleration is \(\mathbf{a}(t)=\langle3,0\rangle\). 2. The rate of change of speed is \(\frac{\mathbf{v}\cdot\mathbf{a}}{\|\mathbf{v}\|}\). 3. At \(t=2\), \(\mathbf{v}(2)=\langle6,4\rangle\), so \(\mathbf{v}\cdot\mathbf{a}=18\) and \(\|\mathbf{v}\|=2\sqrt{13}\). 4. Therefore, the speed changes at \(\frac{9}{\sqrt{13}}\) units per unit time.

\(\frac{9}{\sqrt{13}}\) units per unit time

- The limiting position depends on the total accumulated displacement. - Evaluate the two improper component integrals separately. - Add the resulting displacement vector to the initial position.

1. The total displacement is \(\int_0^{\infty}\mathbf{v}(t)\,dt\). 2. The horizontal displacement is \(\int_0^{\infty}\frac{2}{(1+t)^2}\,dt=2\). 3. The vertical displacement is \(\int_0^{\infty}-3e^{-t}\,dt=-3\). 4. Adding the displacement to \(\langle2,-1\rangle\) gives the limiting position \(\langle4,-4\rangle\).

\(\langle4,-4\rangle\)

- A circle's radius at the particle points in the direction of the position vector. - A tangent direction is perpendicular to that radius. - Use a vector condition to determine the time before evaluating position.

1. The velocity is \(\mathbf{v}(t)=\langle1,2\rangle\). 2. Tangency to the circle centered at the origin requires the position and velocity vectors to be perpendicular. 3. Solve \(\mathbf{r}(t)\cdot\mathbf{v}(t)=(t-1)+2(2t-4)=5t-9=0\), giving \(t=\frac{9}{5}\). 4. The position is \(\mathbf{r}\left(\frac{9}{5}\right)=\left\langle\frac{4}{5},-\frac{2}{5}\right\rangle\).

\(t=\frac{9}{5}\), at position \(\left\langle\frac{4}{5},-\frac{2}{5}\right\rangle\)

- Recover velocity from acceleration and the initial velocity. - Use a dot-product condition for perpendicular vectors. - Apply the positive-time requirement after solving.

1. Integrating the constant acceleration gives \(\mathbf{v}(t)=\langle1+2t,3-t\rangle\). 2. Perpendicularity requires \(\mathbf{v}(t)\cdot\mathbf{a}(t)=0\). 3. The dot product is \(2(1+2t)-(3-t)=5t-1\). 4. Solving \(5t-1=0\) gives \(t=\frac{1}{5}\).

\(t=\frac{1}{5}\)

- Square the speed equation to remove the radical. - Express the squared speed as the sum of the squared velocity components. - Apply the time-domain restriction after solving the quadratic.

1. The squared speed is \(t^2+(t+2)^2\). 2. Set this equal to \((2\sqrt{2})^2=8\): \(2t^2+4t+4=8\). 3. The resulting equation is \(t^2+2t-2=0\), with solutions \(t=-1\pm\sqrt{3}\). 4. The nonnegative solution is \(t=\sqrt{3}-1\).

\(t=\sqrt{3}-1\)

- Differentiate the position twice. - Compare the acceleration vector directly with the position vector. - Use trigonometric identities to simplify both magnitudes.

1. The velocity is \(\mathbf{v}(t)=\langle-12\sin(3t),12\cos(3t)\rangle\), so the speed is \(12\). 2. The acceleration is \(\mathbf{a}(t)=\langle-36\cos(3t),-36\sin(3t)\rangle\). 3. Since \(\mathbf{a}(t)=-9\mathbf{r}(t)\), acceleration points opposite the position vector and therefore toward the origin. 4. Its magnitude is \(36\).

\(\mathbf{a}(t)=-9\mathbf{r}(t)\), so acceleration always points toward the origin. The speed is \(12\), and the acceleration magnitude is \(36\).

- How does the integral of acceleration relate to the change in velocity? - Apply Simpson’s rule separately to the two component graphs. - The result is a change in velocity, not the final velocity itself.

1. The time step is \(h=1\). Simpson’s rule gives a horizontal velocity change of approximately \(\frac{1}{3}\left(0+4(2)+4\right)=4\). 2. The vertical velocity change is approximately \(\frac{1}{3}\left(1+4(0)+1\right)=\frac{2}{3}\). 3. Add both changes to the initial velocity: \(\mathbf{v}(2)\approx\langle1,0\rangle+\left\langle4,\frac{2}{3}\right\rangle=\left\langle5,\frac{2}{3}\right\rangle\).

\(\mathbf{v}(2)\approx\left\langle5,\frac{2}{3}\right\rangle\)

- First express speed as the magnitude of the velocity vector. - Use a variable upper limit for accumulated distance. - Solve the resulting equation subject to the time restriction.

1. The speed is \(\sqrt{(2t)^2+(t^2-1)^2}=\sqrt{(t^2+1)^2}=t^2+1\). 2. The distance traveled by time \(T\) is \(\int_0^T(t^2+1)\,dt=\frac{T^3}{3}+T\). 3. Set \(\frac{T^3}{3}+T=\frac{4}{3}\), which gives \(T^3+3T-4=0\). 4. Factor to obtain \((T-1)(T^2+T+4)=0\). The nonnegative real solution is \(T=1\).

\(t=1\)

- Average velocity uses displacement, while average speed uses distance. - Compute the endpoint positions before finding displacement. - The velocity magnitude is constant on this path.

1. The displacement is \(\mathbf{r}(\pi)-\mathbf{r}(0)=\langle-2,0\rangle\). 2. Dividing by the elapsed time \(\pi\) gives the average velocity \(\left\langle-\frac{2}{\pi},0\right\rangle\). 3. The speed is constantly \(1\), so the distance traveled is \(\pi\). 4. Dividing distance by elapsed time gives average speed \(1\).

Average velocity: \(\left\langle-\frac{2}{\pi},0\right\rangle\); average speed: \(1\)

- A collision requires both coordinate equations to hold at the same time. - After finding the common time, evaluate each position to confirm it. - Differentiate both position functions and compare the resulting vectors.

1. Equal x-coordinates require \(t=2-t\), so \(t=1\). 2. At \(t=1\), both y-coordinates equal \(1\), so the particles collide at \((1,1)\). 3. Their velocities are \(\mathbf{v}_1(t)=\langle1,2t\rangle\) and \(\mathbf{v}_2(t)=\langle-1,-2t\rangle\). 4. At the collision, \(\mathbf{v}_1(1)=\langle1,2\rangle\) and \(\mathbf{v}_2(1)=\langle-1,-2\rangle\), so the velocity vectors are opposites.

They collide at \(t=1\) at \((1,1)\). Their velocity vectors are \(\langle1,2\rangle\) and \(\langle-1,-2\rangle\), which are opposites.

- Interpret “horizontal velocity” as a condition on the two components of the velocity vector. - After finding the qualifying time, use the original position function to locate the particle. - Separately verify the maximum claim over the full stated time interval.

1. The velocity is \(\mathbf v(t)=\langle2t-2,-2t+6\rangle\). 2. Horizontal velocity requires the vertical component to be zero while the horizontal component is nonzero. Thus \(-2t+6=0\), so \(t=3\), and \(\mathbf v(3)=\langle4,0\rangle\). 3. The position is \(\mathbf r(3)=\langle3,10\rangle\). 4. The vertical coordinate \(-t^2+6t+1\) is a downward-opening quadratic with its vertex at \(t=3\), so \(10\) is the maximum y-coordinate on \([0,6]\).

At \(t=3\), the velocity is \(\langle4,0\rangle\) and the position is \((3,10)\). The y-coordinate \(10\) is the maximum on \([0,6]\).

- Both position coordinates must be zero at the same parameter value. - Check all solutions within the stated interval. - Differentiate the position and evaluate at each visit.

1. The x-coordinate is \(0\) when \(t=\pm1\), and both values also make \(t^3-t=0\). 2. Therefore, the particle visits the origin at \(t=-1\) and \(t=1\). 3. The velocity is \(\mathbf{v}(t)=\langle2t,3t^2-1\rangle\). 4. The two velocity vectors are \(\mathbf{v}(-1)=\langle-2,2\rangle\) and \(\mathbf{v}(1)=\langle2,2\rangle\).

The particle is at the origin at \(t=-1\) and \(t=1\). Its velocity vectors are \(\langle-2,2\rangle\) and \(\langle2,2\rangle\), respectively.

- Form the separation vector before differentiating the distance. - Use the relative velocity, which is the derivative of the separation vector. - The sign of the final scalar rate indicates whether the particles are separating or approaching.

1. The separation vector is \(\mathbf{d}(t)=\mathbf{r}_1(t)-\mathbf{r}_2(t)=\langle-t,t^2+t-4\rangle\). 2. At \(t=1\), \(\mathbf{d}(1)=\langle-1,-2\rangle\), and \(\mathbf{d}'(1)=\langle-1,3\rangle\). 3. If \(D=\|\mathbf{d}\|\), then \(D'=\frac{\mathbf{d}\cdot\mathbf{d}'}{\|\mathbf{d}\|}\). 4. Thus \(D'(1)=\frac{(-1)(-1)+(-2)(3)}{\sqrt{5}}=-\sqrt{5}\).

The distance is decreasing at \(\sqrt{5}\) units per unit time; equivalently, \(D'(1)=-\sqrt{5}\).

- The dot product of position and velocity tells you whether distance is increasing or decreasing away from the origin. - Factor the resulting expression and check its sign on the stated interval. - At which time does the point actually reach the origin?

1. Differentiate to get \(\mathbf{v}(t)=\langle1,2t\rangle\). 2. Away from the origin, the sign of \(\mathbf{r}(t)\cdot\mathbf{v}(t)\) determines whether the distance is increasing or decreasing. 3. Compute \(\mathbf{r}(t)\cdot\mathbf{v}(t)=(t-2)(2t^{2}+4t+1)\). The quadratic factor is positive for \(t\ge0\). 4. The distance decreases on \(0\le t<2\) and increases on \(2<t\le4\). At \(t=2\), the point passes through the origin and the distance has a sharp minimum.

Decreasing: \(0\le t<2\). Increasing: \(2<t\le4\). At \(t=2\), the distance is \(0\) and has a sharp minimum.