Free Math Worksheets

- What condition does the first derivative satisfy at the highest point? - How does the second derivative confirm that the point is a maximum? - What does the value of \(h(t)\) represent?

1. Differentiate: \(h'(t) = -9.8t + 12\) and \(h''(t) = -9.8\). 2. Solve \(h'(t) = 0\): \(-9.8t + 12 = 0\), so \(t = \frac{60}{49} \approx 1.224\,\text{s}\). 3. Since \(h''(t) = -9.8 < 0\), the critical time gives a maximum. 4. Evaluate the height: \(h(\frac{60}{49}) = \frac{867}{98} \approx 8.85\,\text{m}\).

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.

- How are height and velocity related? - What is the vertical velocity at the highest point? - Which operation gives the rate of change of a function?

1. Differentiate height: \(v(t) = h'(t) = -10t + 20\). 2. Evaluate \(v(1) = -10(1) + 20 = 10\,\text{m/s}\). 3. At the highest point, \(v(t) = 0\). Solve \(-10t + 20 = 0\) to get \(t = 2\,\text{s}\). 4. Evaluate the height: \(h(2) = -5(2^2) + 20(2) + 1.2 = 21.2\,\text{m}\).

a) \(10\,\text{m/s}\) b) The ball reaches its highest point at \(t = 2\,\text{s}\), at a height of \(21.2\,\text{m}\).

- 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 \(4800\,\text{N}\).

- 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.

- 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. - To maximize velocity, examine the derivative of the velocity function. - Confirm that the critical time lies within the first \(10\) seconds. - Use \(5280\,\text{ft} = 1\,\text{mi}\) and \(3600\,\text{s} = 1\,\text{h}\).

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 \(a(t) = v'(t) = -3t + 18\). Set \(a(t) = 0\): \(-3t + 18 = 0\), so \(t = 6\,\text{s}\). Since \(v''(t) = -3 < 0\), the velocity is maximized then. 4. Evaluate \(v(6) = -1.5(6^2) + 18(6) = 54\,\text{ft/s}\). 5. Convert units: \(54 \cdot \frac{3600}{5280} \approx 36.82\,\text{mph}\).

a) \(v(t) = -1.5t^2 + 18t\); \(v(4) = 48\,\text{ft/s}\) b) \(t = 6\,\text{s}\) c) \(54\,\text{ft/s} \approx 36.82\,\text{mph}\)

- 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. - Total distance is the integral of speed; first check whether velocity changes sign. - Apply the power rule carefully when integrating.

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. The acceleration is \(a(20)=-3\,\text{m/s}^2\), so the velocity is decreasing most rapidly then; the deceleration magnitude is \(3\,\text{m/s}^2\). 3. The velocity factors as \(v(t)=0.01t(t-30)^2\), so it is nonnegative throughout the interval. Thus total distance equals \(\int_0^{30}v(t)\,dt\). An antiderivative is \(0.0025t^4-0.2t^3+4.5t^2\). Evaluation gives \(675\,\text{m}\).

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 \(3\,\text{m/s}^2\). c) \(675\,\text{m}\).

- Acceleration is the derivative of speed with respect to time. - Maximum speed occurs at the highest point of the speed graph. - Maximum acceleration occurs where the graph rises most steeply. - The sign of the slope indicates whether speed is increasing or decreasing.

1. The highest point on the speed graph occurs at \(t=20\,\text{s}\), where \(v=10\,\text{m/s}\). 2. Acceleration is the tangent slope of the speed graph. At \(t=10\,\text{s}\), the slope is approximately \(0.75\,\text{m/s}^2\). 3. Acceleration is greatest where the speed graph rises most steeply. This occurs at the inflection point \(t=10\,\text{s}\). 4. At \(t=5\,\text{s}\), the slope is approximately \(0.56\,\text{m/s}^2\), so the bicyclist is speeding up. At \(t=25\,\text{s}\), the slope is approximately \(-0.94\,\text{m/s}^2\), so 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 \(t=10\,\text{s}\) d) At \(5\,\text{s}\), the rate is positive and the bicyclist is speeding up. At \(25\,\text{s}\), the rate is negative and the bicyclist is slowing down.

- Acceleration is the derivative of speed. - Find times when the speed graph has horizontal tangent lines. - Determine whether speed is increasing or decreasing at the beginning. - Match both the zeros and the sign pattern.

1. Acceleration is the derivative of the speed function: \(a(t)=v^{\prime}(t)\). 2. The speed 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 speed 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 \(a(t)\)

- Recall the derivative and integral relationships among position, velocity, and acceleration. - Use the first derivative of velocity to locate its maximum. - Analyze the acceleration function to find its maximum. - Average velocity can be found from total displacement divided by total time.

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 \(a(4)=2.4\,\text{m/s}^2\). 3. Velocity is nonnegative on \([0,12]\), so total distance is \(\int_0^{12}v(t)\,\text{d}t=[-0.0125t^4+0.2t^3]_0^{12}=86.4\,\text{m}\). 4. The average velocity is \(\frac{1}{12}\int_0^{12}v(t)\,\text{d}t=\frac{86.4}{12}=7.2\,\text{m/s}\).

a) At \(t=8\,\text{s}\); \(12.8\,\text{m/s}\) b) At \(t=4\,\text{s}\); \(2.4\,\text{m/s}^2\) c) \(86.4\,\text{m}\) d) \(7.2\,\text{m/s}\)

- 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 \(h(10)\approx 45\).

- Recall the relationships among position, velocity, and acceleration. - What condition must hold at an interior point where acceleration has a maximum? - How do you evaluate a definite integral when an antiderivative is given? - Include appropriate units in each 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 \(t=20\pm10\sqrt{2}\). Comparing \(a(t)\) at these times and at the endpoints \(t=0\) and \(t=60\) shows that the maximum occurs at \(t=20-10\sqrt{2}\approx5.86\), where \(a(t)\approx3.69\,\text{m/s}^2\). 3. The distance traveled during the first \(20\) seconds is \(\int_0^{20}v(t)\,\text{d}t=V(20)-V(0)=-8000e^{-2}+1600\approx517.32\,\text{m}\).

1. Approximately \(3.69\,\text{m/s}^2\) 2. Approximately \(517.32\,\text{m}\)