Free math worksheets

- Velocity includes a sign; speed does not. - Speed is the magnitude of velocity. - Use the sign of velocity to determine direction.

1. The velocity is already given as \(-7\,\text{m/s}\). 2. Speed is the magnitude of velocity, so the speed is \(|-7|=7\,\text{m/s}\). 3. The negative velocity means the cart is moving in the negative direction along the track.

Velocity: \(-7\,\text{m/s}\); speed: \(7\,\text{m/s}\); direction: negative direction along the track.

- Direction comes from the sign of velocity. - Being at rest at one instant is not enough by itself to prove a direction change. - Compare the velocity signs on the two sides of \(t=3\).

1. Positive velocity on \(0<t<3\) means the particle moves in the positive direction. 2. Negative velocity on \(3<t<8\) means the particle moves in the negative direction. 3. Because the velocity changes sign from positive to negative at \(t=3\), the particle changes direction there.

The particle moves in the positive direction before \(t=3\), moves in the negative direction after \(t=3\), and changes direction at \(t=3\).

- What does the sign of vertical velocity tell you about whether the ball is moving upward or downward? - Find the time when the vertical velocity is zero. - Check how the sign of the velocity changes around that time before evaluating the height.

1. Differentiate height to get vertical velocity: \(v(t)=h'(t)=-9.8t+12\). 2. Solve \(v(t)=0\): \(-9.8t+12=0\), so \(t=\frac{60}{49}\approx 1.224\,\text{s}\). 3. For times just before \(\frac{60}{49}\), \(v(t)>0\), so the ball is moving upward. For times just after \(\frac{60}{49}\), \(v(t)<0\), so it is moving downward. Therefore, the ball is at its highest point at that time. 4. Evaluate the height: \(h\left(\frac{60}{49}\right)=\frac{867}{98}\approx 8.85\,\text{m}\).

The ball reaches its highest point after approximately \(1.22\,\text{s}\), at a height of approximately \(8.85\,\text{m}\).

- Which physical quantity is the first derivative of position? - How are velocity and acceleration related by derivatives? - Substitute the given time into the appropriate function to find an instantaneous value. - To find when acceleration has a specified value, set up and solve an equation in \(t\).

1. Velocity is the derivative of position: \(v(t)=s'(t)=1.5t^2-6t+10\). Acceleration is the derivative of velocity: \(a(t)=v'(t)=3t-6\). 2. Evaluate the velocity: \(v(2)=1.5(2)^2-6(2)+10=4\,\text{m/s}\). 3. Set acceleration equal to zero: \(3t-6=0\), so \(t=2\,\text{s}\).

1) \(v(t)=1.5t^2-6t+10\) and \(a(t)=3t-6\) 2) \(4\,\text{m/s}\) 3) \(t=2\,\text{s}\)

- The first derivative of position gives velocity. - Acceleration is the derivative of velocity. - A negative acceleration does not by itself mean slowing down; compare its sign with the velocity's sign. - Include the correct units for velocity and acceleration.

1. Differentiate position to find velocity: \(v(t)=s'(t)=0.06t^2-0.72t+2.4\). 2. Evaluate: \(v(5)=0.06(5)^2-0.72(5)+2.4=0.3\,\text{m/s}\). 3. Differentiate velocity to find acceleration: \(a(t)=v'(t)=0.12t-0.72\). 4. Evaluate: \(a(5)=0.12(5)-0.72=-0.12\,\text{m/s}^2\). 5. Because the velocity is positive and the acceleration is negative, the robot is slowing down at that instant.

a) \(v(5)=0.3\,\text{m/s}\) b) \(a(5)=-0.12\,\text{m/s}^2\). Since velocity and acceleration have opposite signs, the robot is slowing down.

- Acceleration is the second derivative of position. - Find the acceleration at the specified time before applying \(F=ma\). - Check that the units combine to give newtons.

1. Differentiate position to find velocity: \(v(t)=s'(t)=\frac{1}{4}t^2+2t+5\). 2. Differentiate again to find acceleration: \(a(t)=s''(t)=\frac{1}{2}t+2\). 3. Evaluate: \(a(4)=4\,\text{m/s}^2\). 4. Apply Newton's second law: \(F=ma=(1200\,\text{kg})(4\,\text{m/s}^2)=4800\,\text{N}\).

The net force is \(4800\,\text{N}\).

- Speed is \(|v|\), so ask whether acceleration pushes velocity farther from zero or toward zero. - Compare the signs of \(v\) and \(a\) at each instant. - Negative acceleration does not automatically mean slowing down.

1. A particle speeds up when velocity and acceleration have the same sign, because acceleration changes velocity away from zero. 2. At A, both signs are positive, so the particle is speeding up. At C, both signs are negative, so the particle is also speeding up. 3. A particle slows down when velocity and acceleration have opposite signs. Therefore, it slows down at B and D.

Speeding up: A and C. Slowing down: B and D. The particle speeds up when \(v\) and \(a\) have the same sign and slows down when they have opposite signs.

- Read each quantity from its own table column; position, velocity, and acceleration are different quantities. - Convert velocity to speed by taking its magnitude. - To classify speeding up or slowing down, compare the signs of velocity and acceleration rather than using acceleration alone.

1. At \(t=2\), the table gives \(s(2)=10\,\text{m}\), \(v(2)=-4\,\text{m/s}\), and \(a(2)=-2\,\text{m/s}^2\). 2. The speed is \(|v(2)|=4\,\text{m/s}\), and the negative velocity means the robot is moving in the negative direction. 3. Since velocity and acceleration are both negative, the magnitude of velocity is increasing, so the robot is speeding up. 4. A negative acceleration means velocity is becoming more negative; whether that increases or decreases speed depends on the sign of velocity.

a) Position: \(10\,\text{m}\); velocity: \(-4\,\text{m/s}\); speed: \(4\,\text{m/s}\); direction: negative direction. b) The robot is speeding up because \(v(2)\) and \(a(2)\) are both negative. Negative acceleration alone does not determine whether speed is increasing or decreasing.

- How are position and velocity related? - What is the velocity when the rocket changes direction? - What equation describes the moment when the rocket reaches the ground? - What does a negative velocity mean in this context?

1. Velocity is the derivative of position: \(v(t) = h'(t) = -10t + 30\). 2. At the highest point, \(v(t) = 0\). Solve \(-10t + 30 = 0\) to get \(t = 3\,\text{s}\). Then \(h(3) = -5(3^2) + 30(3) + 0.5 = 45.5\,\text{m}\). 3. The rocket hits the ground when \(h(t) = 0\). The positive solution of \(-5t^2 + 30t + 0.5 = 0\) is \(t = 3 + \frac{\sqrt{910}}{10} \approx 6.017\,\text{s}\). 4. Evaluate the velocity at that time: \(v(t) = -10t + 30 = -\sqrt{910} \approx -30.17\,\text{m/s}\). The negative sign indicates downward motion.

a) \(v(t) = -10t + 30\) b) The rocket reaches its highest point at \(t = 3\,\text{s}\), and the maximum height is \(45.5\,\text{m}\). c) The impact velocity is approximately \(-30.17\,\text{m/s}\), or \(30.17\,\text{m/s}\) downward.

- Average vertical velocity uses the change in height over the change in time. - The sign of \(v(t)=h'(t)\) tells whether the balloon is rising or falling. - A change from positive velocity to negative velocity identifies when upward motion ends. - Acceleration describes how velocity is changing; interpret its sign in terms of velocity, not height.

1. Since \(h(0)=20\) and \(h(5)=70\), the average vertical velocity on \([0,5]\) is \(\frac{70-20}{5}=10\,\frac{\text{m}}{\text{min}}\). 2. The velocity is \(v(t)=h'(t)=-0.6t^2+6t=-0.6t(t-10)\). Thus \(v(3)=12.6\,\frac{\text{m}}{\text{min}}\). 3. On \(0<t<10\), \(v(t)>0\), and on \(10<t\le 15\), \(v(t)<0\). Therefore, the balloon stops rising and begins falling at \(t=10\,\text{min}\). Its height then is \(h(10)=120\,\text{m}\). 4. The acceleration is \(a(t)=v'(t)=-1.2t+6\). Thus \(a(3)=2.4\,\frac{\text{m}}{\text{min}^2}\) and \(a(7)=-2.4\,\frac{\text{m}}{\text{min}^2}\). At \(t=3\), the positive acceleration means the vertical velocity is increasing; at \(t=7\), the negative acceleration means the vertical velocity is decreasing.

a) \(10\,\frac{\text{m}}{\text{min}}\) b) \(12.6\,\frac{\text{m}}{\text{min}}\) c) The balloon stops rising and begins falling at \(t=10\,\text{min}\), when its height is \(120\,\text{m}\). d) \(a(3)=2.4\,\frac{\text{m}}{\text{min}^2}\), so the vertical velocity is increasing then. \(a(7)=-2.4\,\frac{\text{m}}{\text{min}^2}\), so the vertical velocity is decreasing then.

- Acceleration is the derivative of velocity. - Find the absolute maximum of velocity by checking critical values and endpoints. - At an inflection point of velocity, interpret the tangent slope as acceleration with the correct units.

1. The first two derivatives are \(v'(t)=-0.3t^2+3.6t\) and \(v''(t)=-0.6t+3.6\). 2. Solve \(v'(t)=0\): \(t=0\) or \(t=12\). Since \(v''(12)=-3.6<0\), \(t=12\) gives a local maximum. Comparing \(v(0)=5\), \(v(12)=91.4\), and \(v(15)=72.5\) confirms that the maximum velocity occurs after \(12\) seconds. 3. Solve \(v''(t)=0\): \(t=6\). Because \(v'''(t)=-0.6<0\), the acceleration \(v'(t)\) reaches its maximum there. The slope is \(v'(6)=10.8\,\frac{\text{m}}{\text{s}^2}\). Thus, after \(6\) seconds, the vehicle has its greatest acceleration.

a) After \(12\) seconds b) The inflection time is \(t=6\) seconds. The tangent slope is \(10.8\,\frac{\text{m}}{\text{s}^2}\), the vehicle's maximum acceleration.

- What physical quantity does the derivative of height represent? - What is the vertical velocity at the instant the arrow changes direction? - How does the sign of the first derivative indicate whether the height is increasing? - The motion begins at \(t = 0\).

1. Differentiate: \(h'(t) = -9.81t + 45\). 2. Set the velocity equal to zero: \(-9.81t + 45 = 0\), so \(t = \frac{500}{109} \approx 4.587\,\text{s}\). 3. Evaluate the height: \(h(\frac{500}{109}) = \frac{11250}{109} \approx 103.21\,\text{m}\). 4. The arrow is rising when \(h'(t) > 0\). Solving \(-9.81t + 45 > 0\) gives \(0 \le t < \frac{500}{109}\). Therefore, the arrow rises on \([0, \frac{500}{109})\) seconds, approximately \([0, 4.587)\) seconds.

a) \(t = \frac{500}{109} \approx 4.59\,\text{s}\) b) \(\frac{11250}{109}\,\text{m} \approx 103.21\,\text{m}\) c) \([0, \frac{500}{109})\) seconds, approximately \([0, 4.59)\) seconds

- The slope of a position function is instantaneous velocity. - Velocity is the rate of change of position. - Acceleration is the rate of change of velocity. - Equal accelerations can be found by setting the acceleration functions equal.

1. Differentiate to find velocity: \(v_1(t)=12t+2\) and \(v_2(t)=3t^2+6t+5\). At \(t=0\), \(v_1(0)=2\,\text{cm/s}\) and \(v_2(0)=5\,\text{cm/s}\), so car 2 has the greater initial velocity. 2. Differentiate again: \(a_1(t)=12\) and \(a_2(t)=6t+6\). Set them equal: \(12=6t+6\), giving \(t=1\,\text{s}\).

1) Car 2; its velocity is \(5\,\text{cm/s}\), compared with \(2\,\text{cm/s}\) for car 1. 2) \(t=1\,\text{s}\)

- How are position, velocity, and acceleration related by derivatives? - What does instantaneous velocity mean in terms of \(s(t)\)? - At an interior minimum of velocity, what must the acceleration equal? - Track the units through each derivative.

1. Differentiate position to find velocity: \(v(t) = s'(t) = 0.6t^2 - 3t + 4\). 2. Evaluate \(v(5) = 0.6(5^2) - 3(5) + 4 = 4\,\text{m/s}\). 3. Differentiate velocity to find acceleration: \(a(t) = v'(t) = 1.2t - 3\). 4. A minimum of velocity occurs where \(a(t) = 0\). Solve \(1.2t - 3 = 0\) to get \(t = 2.5\,\text{s}\). Because \(v''(t) = 1.2 > 0\), this critical time gives a minimum. 5. At that time, \(a(2.5) = 0\,\text{m/s}^2\).

a) \(4\,\text{m/s}\) b) The velocity is minimized at \(t = 2.5\,\text{s}\), when the acceleration is \(0\,\text{m/s}^2\).

- What must velocity equal when an object is momentarily at rest? - When velocity is zero, its sign on either side determines whether the object reverses direction. - Find rest times by solving a quadratic equation for \(v(t)=0\). - Differentiate position twice to obtain acceleration.

1. Differentiate: \(v(t)=s'(t)=\frac{1}{2}t^2-4t+6\) and \(a(t)=v'(t)=t-4\). 2. At \(t=2\), \(v(2)=0\,\text{m/s}\) and \(a(2)=-2\,\text{m/s}^2\). The object is momentarily at rest and reverses direction because the velocity changes from positive to negative. 3. For another rest time, solve \(v(t)=0\): \(\frac{1}{2}t^2-4t+6=0\). 4. Multiplying by \(2\) and factoring gives \(t^2-8t+12=(t-2)(t-6)=0\). 5. The other time is \(t=6\,\text{s}\), and \(a(6)=2\,\text{m/s}^2\).

a) \(v(2)=0\,\text{m/s}\) and \(a(2)=-2\,\text{m/s}^2\). The object is momentarily at rest and reverses direction. b) The other rest time is \(t=6\,\text{s}\), when \(a(6)=2\,\text{m/s}^2\).

- The object is momentarily at rest when its velocity is zero. - Differentiate velocity to obtain acceleration. - Factor the velocity function and examine its sign on either side of each zero. - A sign change in velocity indicates a reversal of direction.

1. Differentiate position: \(v(t) = s'(t) = 3t^2 - 18t + 24 = 3(t - 2)(t - 4)\). 2. Set \(v(t) = 0\). The object is at rest at \(t = 2\,\text{s}\) and \(t = 4\,\text{s}\). 3. Differentiate velocity: \(a(t) = v'(t) = 6t - 18\). 4. Evaluate: \(a(2) = -6\,\text{m/s}^2\) and \(a(4) = 6\,\text{m/s}^2\). 5. The velocity changes from positive to negative at \(t = 2\), so the object reverses from the positive direction to the negative direction. At \(t = 4\), velocity changes from negative to positive, so the object reverses back to the positive direction.

a) \(t = 2\,\text{s}\) and \(t = 4\,\text{s}\) b) \(a(2) = -6\,\text{m/s}^2\) and \(a(4) = 6\,\text{m/s}^2\) c) At \(t = 2\), motion changes from the positive direction to the negative direction. At \(t = 4\), it changes from the negative direction to the positive direction.

- Use derivatives to relate position, velocity, and acceleration. - To convert \(\text{m/s}\) to \(\text{km/h}\), multiply by \(3.6\). - The slope of a position-time graph is instantaneous velocity.

1. Differentiate position: \(v(t)=s'(t)=0.015t^2+0.12t\). 2. Evaluate: \(v(20)=0.015(20)^2+0.12(20)=8.4\,\text{m/s}\). 3. Convert units: \(8.4\cdot3.6=30.24\), so the velocity is \(30.24\,\text{km/h}\). 4. Differentiate again: \(a(t)=s''(t)=0.03t+0.12\). 5. Evaluate: \(a(20)=0.03(20)+0.12=0.72\,\text{m/s}^2\).

a) \(8.4\,\text{m/s}\), or \(30.24\,\text{km/h}\) b) \(0.72\,\text{m/s}^2\)

- Differentiate position once for velocity and twice for acceleration. - At the requested time, the two velocity functions have equal values. - Discard a negative time because the model uses \(t\geq0\).

1. Differentiate to find velocity: \(v_1(t)=\frac{1}{2}t^2+4t+4\) and \(v_2(t)=2t+10\). 2. Set the velocities equal: \(\frac{1}{2}t^2+4t+4=2t+10\). 3. Simplifying gives \(t^2+4t-12=0=(t-2)(t+6)\). Since \(t\geq0\), the relevant time is \(t=2\,\text{s}\). 4. Differentiate velocity: \(a_1(t)=t+4\) and \(a_2(t)=2\). 5. At \(t=2\), \(a_1(2)=6\,\text{m/s}^2\) and \(a_2(2)=2\,\text{m/s}^2\).

At \(t=2\,\text{s}\), vehicle 1 has acceleration \(6\,\text{m/s}^2\), and vehicle 2 has acceleration \(2\,\text{m/s}^2\).

- The first derivative of position is velocity, and the second derivative is acceleration. - Express equal acceleration by setting the two acceleration functions equal. - Substitute the time from part a) into both velocity functions.

1. Differentiate twice. For object \(A\), \(v_A(t)=\frac{1}{4}t^2+t+2\) and \(a_A(t)=\frac{1}{2}t+1\). For object \(B\), \(v_B(t)=\frac{3}{2}t+3\) and \(a_B(t)=\frac{3}{2}\). 2. Set the accelerations equal: \(\frac{1}{2}t+1=\frac{3}{2}\), so \(t=1\,\text{s}\). 3. Evaluate the velocities: \(v_A(1)=3.25\,\text{m/s}\) and \(v_B(1)=4.5\,\text{m/s}\). 4. Since \(4.5>3.25\), object \(B\) has the greater velocity.

a) \(t=1\,\text{s}\) b) Object \(B\); \(v_B(1)=4.5\,\text{m/s}\), while \(v_A(1)=3.25\,\text{m/s}\).

- Use the chain rule for the square-root function. - Position, velocity, and acceleration are connected by successive derivatives. - Differentiate the velocity using the product rule or quotient rule. - A positive second derivative means the first derivative is increasing.

1. a) \(s(0)=\sqrt{900}=30\), so the initial position is \(30\,\text{m}\) from the origin. 2. b) By the chain rule, \(s'(t)=\frac{4t}{\sqrt{4t^2+900}}\). Therefore, \(s'(20)=\frac{80}{50}=1.6\,\text{m}/\text{s}\). 3. c) Differentiate \(s'(t)=4t(4t^2+900)^{-\frac{1}{2}}\): \(s''(t)=4(4t^2+900)^{-\frac{1}{2}}-16t^2(4t^2+900)^{-\frac{3}{2}}\). 4. Combining terms gives \(s''(t)=\frac{4(4t^2+900)-16t^2}{(4t^2+900)^{\frac{3}{2}}}=\frac{3600}{(4t^2+900)^{\frac{3}{2}}}\). 5. Since \(s''(t)>0\) for \(t\ge0\), the velocity \(s'(t)\) is increasing.

a) \(30\,\text{m}\) b) \(1.6\,\text{m}/\text{s}\) c) \(s''(t)=\frac{3600}{(4t^2+900)^{\frac{3}{2}}}>0\), so the vehicle's velocity is increasing.

- Use the second derivative of position to find acceleration before applying \(F=ma\). - Zero net force means the force function equals \(0\). - Velocity is the first derivative of position. - A negative velocity indicates motion in the negative direction.

1. Differentiate position: \(v(t)=s'(t)=2t^2-12t+10\). 2. Differentiate again: \(a(t)=v'(t)=4t-12\). 3. Apply \(F=ma\): \(F(t)=0.5(4t-12)=2t-6\). 4. Set the force equal to zero: \(2t-6=0\), so \(t=3\,\text{s}\). 5. Evaluate the velocity: \(v(3)=2(3)^2-12(3)+10=-8\,\text{m/s}\).

a) \(F(t)=2t-6\), in newtons b) \(t=3\,\text{s}\) c) \(-8\,\text{m/s}\)

- Average acceleration is the average rate of change of velocity over an interval. - Distinguish an average rate from an instantaneous rate. - Acceleration measures the rate of change of velocity. - Relate velocity to position, then differentiate once more.

1. Compute the endpoint velocities: \(v(2)=6\,\text{m/s}\) and \(v(4)=16\,\text{m/s}\). The average acceleration is \(\frac{v(4)-v(2)}{4-2}=\frac{16-6}{2}=5\,\text{m/s}^2\). 2. Differentiate velocity: \(a(t)=v'(t)=t+2\). Thus \(a(2)=4\,\text{m/s}^2\). 3. Velocity is the first derivative of position, \(v(t)=s'(t)\), and acceleration is the derivative of velocity. Therefore, \(a(t)=s''(t)\).

1) \(5\,\text{m/s}^2\) 2) \(4\,\text{m/s}^2\) 3) \(a(t)=s''(t)\)

- Position, velocity, and acceleration are connected by successive derivatives. - Zero acceleration means the acceleration function equals \(0\). - Use a difference quotient of velocity to find average acceleration. - Consider why a linear acceleration function has its midpoint value equal to its average over a symmetric interval.

1. Differentiate position: \(v(t)=s'(t)=3t^2-12t+12\). 2. Differentiate velocity: \(a(t)=v'(t)=6t-12\). 3. Solve \(6t-12=0\), giving \(t=2\,\text{s}\). 4. The endpoint velocities are \(v(1)=3\,\text{m/s}\) and \(v(3)=3\,\text{m/s}\). Thus the average acceleration is \(\frac{v(3)-v(1)}{3-1}=0\,\text{m/s}^2\). Also, \(a(2)=0\,\text{m/s}^2\), so the two values are equal.

1) \(v(t)=3t^2-12t+12\) 2) \(a(t)=6t-12\) 3) \(t=2\,\text{s}\) 4) The average acceleration is \(0\,\text{m/s}^2\), equal to \(a(2)\).

- Velocity is the derivative of position, and acceleration is the derivative of velocity. - Find where acceleration changes sign within the modeled interval. - What does positive acceleration do to velocity? What does negative acceleration do to velocity? - Use \(5280\,\text{ft}=1\,\text{mi}\) and \(3600\,\text{s}=1\,\text{h}\) for the final conversion.

1. Differentiate position: \(v(t)=s'(t)=-1.5t^2+18t\). 2. Evaluate \(v(4)=-1.5(4^2)+18(4)=48\,\text{ft/s}\). 3. Acceleration is \(a(t)=v'(t)=-3t+18\). Solving \(a(t)=0\) gives \(t=6\,\text{s}\). 4. For \(0\le t<6\), \(a(t)>0\), so velocity is increasing. For \(6<t\le10\), \(a(t)<0\), so velocity is decreasing. Therefore, velocity is greatest at \(t=6\,\text{s}\). 5. Evaluate \(v(6)=54\,\text{ft/s}\). Convert units: \(54\cdot\frac{3600}{5280}\approx36.82\,\text{mph}\).

a) \(v(t)=-1.5t^2+18t\); \(v(4)=48\,\text{ft/s}\) b) \(t=6\,\text{s}\) c) \(54\,\text{ft/s}\approx36.82\,\text{mph}\)

- Find critical times of the velocity and compare them with the interval endpoints. - Acceleration is the derivative of velocity. - The most rapid decrease occurs where acceleration is smallest. - Compare any acceleration critical point with the interval endpoints if needed.

1. Differentiate: \(v'(t)=0.03t^2-1.2t+9\). The critical times are \(t=10\) and \(t=30\). Compare the velocity at the endpoints and the interior critical time: \(v(0)=0\), \(v(10)=40\), and \(v(30)=0\). Therefore, the maximum velocity is \(40\,\text{m/s}\) at \(t=10\,\text{s}\). 2. Acceleration is \(a(t)=v'(t)\). Its derivative is \(a'(t)=v''(t)=0.06t-1.2\), which is zero at \(t=20\). Since acceleration is an upward-opening quadratic, this is its minimum. 3. The acceleration is \(a(20)=-3\,\text{m/s}^2\), so the velocity is decreasing most rapidly then; the deceleration magnitude is \(3\,\text{m/s}^2\).

a) \(40\,\text{m/s}\) at \(t=10\,\text{s}\). b) At \(t=20\,\text{s}\), the acceleration is \(-3\,\text{m/s}^2\), corresponding to a deceleration magnitude of \(3\,\text{m/s}^2\).

- Acceleration is the derivative of signed velocity with respect to time. - Because the displayed velocity never becomes negative, its magnitude is also the speed. - Estimate acceleration from the tangent slope of the velocity graph at the requested time. - To decide speeding up or slowing down, compare the signs of velocity and acceleration.

1. The highest point on the velocity graph occurs at \(t=20\,\text{s}\), where \(v=10\,\text{m/s}\). Since the velocity is nonnegative, this is also the maximum speed. 2. Acceleration is the tangent slope of the velocity graph. At \(t=10\,\text{s}\), the slope is approximately \(0.75\,\text{m/s}^2\). 3. At \(t=5\,\text{s}\), the slope is approximately \(0.56\,\text{m/s}^2\). The positive acceleration and positive velocity mean the bicyclist is speeding up. 4. At \(t=25\,\text{s}\), the slope is approximately \(-0.94\,\text{m/s}^2\). The negative acceleration and positive velocity mean the bicyclist is slowing down.

a) At \(t=20\,\text{s}\), with a maximum speed of \(10\,\text{m/s}\) b) Approximately \(0.75\,\text{m/s}^2\) c) At \(5\,\text{s}\), \(a\approx0.56\,\text{m/s}^2\), so the bicyclist is speeding up. At \(25\,\text{s}\), \(a\approx-0.94\,\text{m/s}^2\), so the bicyclist is slowing down.

- Acceleration is the derivative of signed velocity. - Find times when the velocity graph has horizontal tangent lines. - Determine whether the velocity is increasing or decreasing at the beginning. - A correct derivative graph must match both the zero locations and the sign pattern.

1. Acceleration is the derivative of velocity: \(a(t)=v'(t)\). 2. The velocity graph has local maxima at \(t=2\) and \(t=10\) and a local minimum at \(t=6\). Therefore, the acceleration graph must have zeros at \(t=2\), \(t=6\), and \(t=10\). 3. Graph c has zeros at \(t=0\), \(t=4\), \(t=8\), and \(t=12\), so it does not fit. Graphs a and b have the required zeros. 4. At \(t=0\), the velocity is increasing, so \(a(0)>0\). Graph a starts with a positive value, while graph b starts with a negative value. Therefore, graph a represents the acceleration.

Graph a represents the acceleration.

- Velocity is the derivative of position, and acceleration is the derivative of velocity. - At a local maximum or minimum of a function, its derivative is zero. - Compare where one graph increases or decreases with the sign of a possible derivative graph. - Differentiating a cubic produces a quadratic, and differentiating a quadratic produces a linear function.

1. For straight-line motion, \(s^{\prime}(t)=v(t)\) and \(v^{\prime}(t)=a(t)\). 2. Graph \(h\) has a local maximum at \(t=2\) and a local minimum at \(t=6\). Graph \(g\) is zero at those same times, and its sign agrees with when \(h\) increases or decreases. Therefore, \(g=h^{\prime}\), so \(h\) is position and \(g\) is velocity. 3. Graph \(g\) has a local minimum at \(t=4\). Graph \(f\) is zero at \(t=4\), and its sign agrees with when \(g\) decreases or increases. Therefore, \(f=g^{\prime}\), so \(f\) is acceleration.

Graph \(h\): position \(s(t)\); Graph \(g\): velocity \(v(t)\); Graph \(f\): acceleration \(a(t)\)

- Use the derivative of velocity to locate candidates for maximum velocity. - Compare interior candidates with the interval endpoints. - Acceleration is the derivative of velocity. - Analyze the acceleration function to find its maximum.

1. \(v'(t)=-0.15t^2+1.2t=-0.15t(t-8)\). Comparing critical points and endpoints on \([0,12]\) shows that the maximum occurs at \(t=8\), where \(v(8)=12.8\,\text{m/s}\). 2. The acceleration is \(a(t)=-0.15t^2+1.2t\). Since \(a\) is a downward-opening parabola, its maximum occurs at \(t=4\), where \(a(4)=2.4\,\text{m/s}^2\).

a) At \(t=8\,\text{s}\); \(12.8\,\text{m/s}\) b) At \(t=4\,\text{s}\); \(2.4\,\text{m/s}^2\)

- At \(t=10\), move vertically from the time axis to the graph, then read the corresponding altitude. - Find where the graph crosses the horizontal level \(h=40\). - Approximate the slope of a tangent line at \(t=10\). - Interpret each side of the equation as an altitude at a particular time.

1. At \(t=10\), the graph passes through approximately \((10,45)\), so the drone's altitude is about \(45\,\text{m}\). 2. The graph intersects \(h=40\) at approximately \(t=8.1\) and \(t=29.0\). Therefore, the drone is at least \(40\,\text{m}\) high for approximately \(8.1\leq t\leq 29.0\). 3. Estimate the slope of the tangent line at \((10,45)\). A reasonable slope estimate is \(h^{\prime}(10)\approx 2.5\,\text{m/s}\), so the drone is rising at about \(2.5\,\text{m/s}\). 4. The equation means that, starting at time \(t\), the drone is exactly \(15\,\text{m}\) higher \(10\) seconds later. 5. For \(t=10\), the graph gives \(h(10)\approx 45\) and \(h(20)\approx 60\). Since \(60=45+15\), the equation is true when \(t=10\).

a) The altitude after \(10\,\text{s}\) is about \(45\,\text{m}\). The drone is at least \(40\,\text{m}\) high for approximately \(8.1\leq t\leq 29.0\). b) The instantaneous vertical velocity is approximately \(2.5\,\text{m/s}\) upward. c) The equation means that the drone is \(15\,\text{m}\) higher \(10\) seconds after time \(t\). It is true for \(t=10\) because \(h(20)\approx 60\) and \(h(10)\approx 45\).

- Displacement uses only final position minus initial position. - Use the stated velocity signs to know that no hidden reversal occurs inside each interval. - For total distance, add the absolute position changes across the three intervals.

1. Displacement depends only on the final and initial positions: \(s(7)-s(0)=4-2=2\,\text{m}\). 2. Because direction is constant on each stated interval, the distance on \([0,2]\) is \(|7-2|=5\,\text{m}\), on \([2,5]\) it is \(|-1-7|=8\,\text{m}\), and on \([5,7]\) it is \(|4-(-1)|=5\,\text{m}\). 3. Total distance is \(5+8+5=18\,\text{m}\). 4. Displacement records the net change in position, while total distance adds the lengths traveled in each direction.

Displacement: \(2\,\text{m}\). Total distance traveled: \(18\,\text{m}\). They differ because displacement is net position change, while distance counts all motion regardless of direction.

- Recall the relationship between velocity and acceleration. - An interior maximum of acceleration can occur where the derivative of acceleration is zero. - Compare every eligible critical point with the endpoints of the test interval. - Include acceleration units in the final answer.

1. The acceleration is \(a(t)=v'(t)=(1.6t-0.08t^2)e^{-0.1t}\). 2. Critical points of \(a\) satisfy \(a'(t)=v''(t)=(0.008t^2-0.32t+1.6)e^{-0.1t}=0\). This gives \(t=20\pm10\sqrt{2}\). 3. Compare \(a(t)\) at these two critical points and at the endpoints \(t=0\) and \(t=60\). The maximum occurs at \(t=20-10\sqrt{2}\approx5.86\). 4. At that time, \(a(t)\approx3.69\,\text{m/s}^2\).

Approximately \(3.69\,\text{m/s}^2\)