Free math worksheets

- Read both the constant velocity and the elapsed time from the graph. - The region under a constant velocity graph is a rectangle. - Keep the sign of the velocity when interpreting the area as displacement.

1. Displacement is the signed area under the velocity graph. 2. From the graph, the region is a rectangle with height \(6\,\text{m/s}\) and width \(4\,\text{s}\). 3. The displacement is \((6\,\text{m/s})(4\,\text{s})=24\,\text{m}\).

\(24\,\text{m}\)

- Displacement uses signed area under a velocity graph. - How should an area below the time axis affect displacement? - Total distance counts every traveled segment positively.

1. From \(t=0\) to \(t=1\), the signed area is \(2\). From \(t=1\) to \(t=2\), the signed area is \(-2\). 2. The displacement is \(2+(-2)=0\) units. 3. Total distance uses the magnitudes of the two areas, so it is \(2+2=4\) units.

Displacement: \(0\) units Total distance: \(4\) units

- Position changes by the signed area under the velocity graph. - Is the displayed velocity above or below the time axis? - Combine the signed displacement with the given initial position.

1. The graph shows constant velocity \(-2\) from \(t=0\) to \(t=3\), so the displacement is the signed area \((-2)(3)=-6\). 2. Therefore, \(s(3)=s(0)+\text{displacement}=10-6=4\).

\(s(3)=4\)

- How are velocity and distance traveled related? - Which operation reverses differentiation with respect to time? - How can you find the change in position between two times? - Account for the inner derivative when integrating the exponential term.

1. Distance traveled over a time interval is the definite integral of velocity. 2. An antiderivative is \(V(t)=15t+75e^{-0.2t}\). 3. During the first \(5\) seconds, the distance is \(\int_0^5 v(t)\,\text{d}t=[15t+75e^{-0.2t}]_0^5=75e^{-1}\approx27.59\,\text{m}\). 4. During the next \(5\) seconds, the distance is \(\int_5^{10} v(t)\,\text{d}t=[15t+75e^{-0.2t}]_5^{10}=75+75e^{-2}-75e^{-1}\approx57.56\,\text{m}\).

The boat travels approximately \(27.59\,\text{m}\) during the first \(5\,\text{s}\) and approximately \(57.56\,\text{m}\) during the next \(5\,\text{s}\).

- Since the graph stays above the time axis, distance equals the geometric area under the graph. - Break the region into a triangle, a rectangle, and a trapezoid. - Add the three areas and keep track of the units.

1. Because the velocity is nonnegative, the distance is the area under the velocity graph. 2. From \(t=0\) to \(t=4\), the area is a triangle: \(\frac{1}{2}\cdot4\cdot8=16\,\text{m}\). 3. From \(t=4\) to \(t=10\), the area is a rectangle: \(6\cdot8=48\,\text{m}\). 4. From \(t=10\) to \(t=12\), the area is a trapezoid: \(\frac{1}{2}(8+4)\cdot2=12\,\text{m}\). 5. The total distance is \(16+48+12=76\,\text{m}\).

The cyclist travels \(76\,\text{m}\).

- What does the area under a velocity-versus-time graph represent? - How are position and velocity related? - What does the constant of integration represent when an initial position is given? - Recall the power rule for antiderivatives.

1. An antiderivative of \(v(t)=0.6t^2+5\) is \(V(t)=0.2t^3+5t\). 2. \(\int_0^5(0.6t^2+5)\,\text{d}t=[0.2t^3+5t]_0^5=25+25=50\). This is the additional distance, in meters, traveled during the first \(5\) seconds. 3. Since \(s'(t)=v(t)\), the position has the form \(s(t)=0.2t^3+5t+C\). 4. Using \(s(0)=12\) gives \(C=12\), so \(s(t)=0.2t^3+5t+12\).

a) \(50\); the car travels \(50\,\text{m}\) during the first \(5\) seconds. b) \(s(t)=0.2t^3+5t+12\).

- How do you obtain position from velocity? - What does \(s(0)=0\) tell you about the constant of integration? - How can you find the change in a quantity from its rate of change over an interval? - Which calculus tool gives the area under a graph on an interval?

1. Since position is an antiderivative of velocity, \(s(t)=\int 0.75t^2\,\text{d}t=0.25t^3+C\). 2. The condition \(s(0)=0\) gives \(C=0\), so \(s(t)=0.25t^3\). 3. The distance traveled from \(t=2\) to \(t=6\) is \(\int_2^6 0.75t^2\,\text{d}t=[0.25t^3]_2^6=54-2=52\,\text{m}\).

a) \(s(t)=0.25t^3\) b) \(52\,\text{m}\)

- Use the vertical grid lines at the four requested times to read the graph. - Check whether the velocity stays nonnegative before treating area as distance. - For each trapezoid, the two velocity values are the parallel sides and the time interval is the width.

1. Reading the graph gives \(v(0)=2\), \(v(2)=4\), \(v(4)=5\), and \(v(6)=2\), in meters per second. 2. The velocity is nonnegative on the displayed interval, so the distance traveled is the area under the velocity graph. 3. Each trapezoid has width \(2\,\text{s}\). The estimate is \(\frac{2}{2}(2+4)+\frac{2}{2}(4+5)+\frac{2}{2}(5+2)=6+9+7=22\,\text{m}\).

a) \(v(0)=2\,\text{m/s}\), \(v(2)=4\,\text{m/s}\), \(v(4)=5\,\text{m/s}\), \(v(6)=2\,\text{m/s}\) b) \(22\,\text{m}\)

- How are velocity and position related through integration? - Which operation gives distance traveled from a velocity function? - First find an antiderivative that satisfies \(s(0)=0\). - Evaluate the position function at the requested times.

1. Since velocity is nonnegative on \([0,10]\), distance traveled equals the integral of velocity. 2. An antiderivative is \(s(t)=-0.04t^3+0.6t^2+4t+C\). The condition \(s(0)=0\) gives \(C=0\). 3. During the first \(5\) seconds, the distance is \(s(5)-s(0)=-5+15+20=30\,\text{m}\). 4. During the full sprint, the distance is \(s(10)-s(0)=-40+60+40=60\,\text{m}\).

a) \(30\,\text{m}\) b) \(60\,\text{m}\) c) \(s(t)=-0.04t^3+0.6t^2+4t\)

- First decide what area under a velocity-time graph represents when velocity is nonnegative. - Determine the physical quantity represented by one grid square. - Combine partial grid squares rather than counting only the completely filled ones.

1. The displayed velocity is nonnegative, so the distance traveled from \(t=2\) to \(t=8\) is the area under the velocity graph over that interval. 2. Each grid square represents \(1\,\text{s}\cdot1\,\text{m/s}=1\,\text{m}\). 3. Counting whole squares and combining partial squares gives an estimated area of about \(27\) square grid units, so the distance is about \(27\,\text{m}\).

About \(27\,\text{m}\). Whole-meter estimates from \(25\,\text{m}\) to \(29\,\text{m}\) are reasonable from the graph.

- What direction of motion corresponds to a graph above or below the time axis? - The distance from the reference point increases while the car moves in the positive direction. - Divide the signed area into trapezoids and triangles. - Add the net change in position to the initial position of \(50\,\text{m}\).

1. The car moves away from the reference point while \(v(t)>0\), through \(t=8\). After \(t=8\), velocity is negative, so the car moves back toward the reference point. Therefore, its greatest distance occurs at \(t=8\). 2. From \(0\) to \(4\), the signed area is \(\frac{2+6}{2}(4)=16\,\text{m}\). 3. From \(4\) to \(8\), the signed area is \(\frac{1}{2}(4)(6)=12\,\text{m}\). 4. From \(8\) to \(12\), the signed area is \(-\frac{1}{2}(4)(4)=-8\,\text{m}\). 5. The net change in position is \(16+12-8=20\,\text{m}\), so \(s(12)=50+20=70\,\text{m}\).

a) \(t=8\,\text{s}\), because \(v(t)\) changes from positive to negative there. b) \(s(12)=70\,\text{m}\)

- How are velocity and distance traveled related? - Which value of \(t\) corresponds to Kart B's start? - Remember that Kart B does not move before \(t=10\). - Express the distance between the karts using their positions.

1. At \(t=10\), Kart A has traveled \(\int_0^{10}18(1-e^{-0.08t})\,\text{d}t=[18t+225e^{-0.08t}]_0^{10}\approx56.10\,\text{m}\). 2. Ten seconds after Kart B starts corresponds to \(t=20\). The velocities are \(v_A(20)\approx14.37\,\text{m/s}\) and \(v_B(20)\approx16.77\,\text{m/s}\), so the difference is approximately \(2.41\,\text{m/s}\). 3. Twenty seconds after Kart B starts corresponds to \(t=30\). Kart A's position is \(\int_0^{30}v_A(t)\,\text{d}t\approx335.41\,\text{m}\). Kart B's position is \(\int_{10}^{30}v_B(t)\,\text{d}t\approx298.14\,\text{m}\). 4. The distance between the karts is \(335.41-298.14\approx37.27\,\text{m}\).

a) Approximately \(56.10\,\text{m}\) b) Approximately \(2.41\,\text{m/s}\) c) Approximately \(37.27\,\text{m}\)

- What quantity results from integrating velocity over time? - What does a zero difference between the boats' positions mean? - Interpret the lower and upper limits of each integral. - Compare the limiting velocities to determine which boat is eventually faster.

1. The first integral is Boat 1's position relative to the common starting point at time \(x\). The second integral is Boat 2's position at time \(x\). Therefore, \(D(x)\) is the signed position difference: positive values mean Boat 1 is ahead, and negative values mean Boat 2 is ahead. 2. At \(t=50\), Boat 1 has traveled \([15t+300e^{-0.05t}]_0^{50}\approx474.63\,\text{m}\). 3. At \(t=50\), Boat 2 has traveled \([25t+250e^{-0.1(t-25)}]_{25}^{50}\approx395.52\,\text{m}\). 4. Using the unrounded values from Steps 2 and 3, \(D(50)\approx79.10\,\text{m}\), so Boat 1 is approximately \(79.10\,\text{m}\) ahead at \(t=50\). 5. If \(D(z)=0\), the boats have the same position at time \(z\). Since Boat 2 starts later but approaches the greater limiting velocity, such a zero corresponds to Boat 2 catching Boat 1.

a) \(D(x)\) is the signed difference between the boats' positions at time \(x\). b) \(D(50)\approx79.10\,\text{m}\); Boat 1 is approximately \(79.10\,\text{m}\) ahead. c) \(D(z)=0\) means the boats are at the same position, so Boat 2 has caught Boat 1.

- Recall the relationships among position, velocity, and acceleration. - Which rules apply when the variable appears both in a linear factor and in an exponential function? - Distance traveled is found by integrating velocity over time. - What happens to a polynomial factor times \(e^{-0.5t}\) as \(t\) increases without bound?

1. Acceleration is the derivative of velocity. Rewriting \(v(t)=80-(40t+80)e^{-0.5t}\) and applying the product and chain rules gives \(a(t)=20te^{-0.5t}\). 2. The distance traveled is \(\int_0^{20}v(t)\,\text{d}t\). 3. An antiderivative is \(V(t)=80t+(80t+320)e^{-0.5t}\). 4. Therefore, \(\int_0^{20}v(t)\,\text{d}t=[80t+(80t+320)e^{-0.5t}]_0^{20}=1280+1920e^{-10}\approx1280.09\,\text{m}\). 5. Since \((0.5t+1)e^{-0.5t}\to0\) as \(t\to\infty\), \(v(t)\to80\,\text{m/s}\).

a) \(a(t)=20te^{-0.5t}\) b) Approximately \(1280.09\,\text{m}\) c) \(80\,\text{m/s}\)

- Recall the formula for integration by parts. - How do you use an antiderivative to find distance over a time interval? - What does the average value of a continuous velocity function represent? - Check the signs carefully when evaluating the antiderivative.

1. Integration by parts gives \(V(t)=(-50t-500)e^{-0.2t}\). 2. The distance traveled is \(\int_5^{15}v(t)\,\text{d}t=V(15)-V(5)=-1250e^{-3}+750e^{-1}\approx213.68\,\text{m}\). 3. The average velocity is \(\frac{1}{15-5}\int_5^{15}v(t)\,\text{d}t\approx\frac{213.68}{10}=21.37\,\text{m/s}\).

1. \(V(t)=(-50t-500)e^{-0.2t}\) 2. Approximately \(213.68\,\text{m}\) 3. Approximately \(21.37\,\text{m/s}\)

- How are velocity and position related? - What does a known point on the position function tell you about \(C\)? - Recall the fundamental theorem of calculus. - What happens to the same constant when two values of an antiderivative are subtracted?

1. Integrating velocity gives \(s(t)=0.1t^3+2t+C\). 2. Using \(s(1)=5\) gives \(0.1+2+C=5\), so \(C=2.9\). Thus, \(s(t)=0.1t^3+2t+2.9\). 3. \(s(4)=17.3\), so \(s(4)-s(1)=17.3-5=12.3\,\text{m}\). 4. Also, \(\int_1^4(0.3t^2+2)\,\text{d}t=[0.1t^3+2t]_1^4=12.3\,\text{m}\). 5. The results agree by the fundamental theorem of calculus: \(s(b)-s(a)=\int_a^bv(t)\,\text{d}t\). The constant of integration cancels in the difference.

1. \(s(t)=0.1t^3+2t+C\) 2. \(C=2.9\), so \(s(t)=0.1t^3+2t+2.9\). 3. a) \(12.3\,\text{m}\) b) \(12.3\,\text{m}\) 4. The values are equal because the definite integral of velocity gives the change in position.

- How do you integrate an exponential function of the form \(e^{kt}\)? - What happens to \(e^{-0.5t}\) as \(t\) increases without bound? - In part 3, write the position function as an antiderivative plus a constant.

1. \(\int_0^4 12e^{-0.5t}\,\text{d}t=[-24e^{-0.5t}]_0^4=24(1-e^{-2})\,\text{m}\). 2. The total distance is the improper integral \(\int_0^\infty12e^{-0.5t}\,\text{d}t=\lim_{b\to\infty}[-24e^{-0.5t}]_0^b=24\,\text{m}\). The integral converges, so the total distance is bounded. 3. If \(S\) is an antiderivative of \(v\), then any position function has the form \(s(t)=S(t)+C\). Therefore, \(s(b)-s(a)=(S(b)+C)-(S(a)+C)=S(b)-S(a)\), so the initial-position constant cancels.

1. \(24(1-e^{-2})\,\text{m}\), which is approximately \(20.75\,\text{m}\) 2. The improper integral converges to \(24\,\text{m}\), so the total distance is finite. 3. The constant that determines initial position cancels from \(s(b)-s(a)\).

- Velocity is an antiderivative of acceleration, and position is an antiderivative of velocity. - Rewrite the square root as a power before integrating. - Use each initial condition to determine the constant of integration. - To evaluate a power such as \(16^{5/2}\), take the square root first.

1. Integrating acceleration gives \(v(t)=\int 0.75t^{1/2}\,\text{d}t=0.5t^{3/2}+C_1\). 2. Since \(v(0)=0\), \(C_1=0\), so \(v(t)=0.5t^{3/2}\). 3. Integrating velocity gives \(s(t)=\int 0.5t^{3/2}\,\text{d}t=0.2t^{5/2}+C_2\). 4. Since \(s(0)=0\), \(C_2=0\), so \(s(t)=0.2t^{5/2}\). 5. \(s(16)=0.2(16)^{5/2}=0.2(4^5)=204.8\,\text{m}\).

a) \(v(t)=0.5t^{3/2}\) b) \(s(t)=0.2t^{5/2}\) c) \(204.8\,\text{m}\)

- Integrate the exponential acceleration. - Use the initial velocity to determine the first constant. - Integrate again and use the initial position. - Check the sign of velocity before equating displacement and distance.

1. Integrating acceleration gives \(v(t)=8e^{-0.5t}+C\). Since \(v(0)=8\), \(C=0\), so \(v(t)=8e^{-0.5t}\). 2. Integrating velocity gives \(s(t)=-16e^{-0.5t}+C\). Since \(s(0)=0\), \(C=16\). 3. Thus, \(s(t)=16-16e^{-0.5t}\). 4. Since the velocity is positive, the distance traveled in the first \(2\) seconds is \(s(2)-s(0)=16(1-e^{-1})\approx10.11\,\text{m}\).

a) \(v(t)=8e^{-0.5t}\) b) \(s(t)=16-16e^{-0.5t}\) c) \(16(1-e^{-1})\,\text{m}\approx10.11\,\text{m}\)

- Integrate acceleration, then velocity. - Apply the initial conditions. - Set the position function equal to \(160\). - Reject any negative-time solution.

1. Integrating acceleration gives \(v(t)=2t+C_1\). Since \(v(0)=5\), \(v(t)=2t+5\). 2. Integrating velocity gives \(s(t)=t^2+5t+C_2\). Since \(s(0)=10\), \(s(t)=t^2+5t+10\). 3. Set \(t^2+5t+10=160\), giving \(t^2+5t-150=0\). 4. The roots are \(t=10\) and \(t=-15\). The physical solution is \(t=10\,\text{s}\).

a) \(v(t)=2t+5\) and \(s(t)=t^2+5t+10\) b) \(t=10\,\text{s}\)

- Relate positive, zero, and negative velocity to changes in altitude. - Use signed area under the velocity graph to find changes in altitude. - Break the regions into triangles and rectangles.

1. The altitude is constant when \(v(t)=0\), which occurs on \(6\le t\le7\). 2. The balloon rises until \(t=6\), remains at the same altitude until \(t=7\), and then descends. Therefore, it is at its maximum altitude from \(t=6\) to \(t=7\). 3. The signed area above the axis from \(t=0\) to \(t=6\) is \(\frac{1}{2}\cdot2\cdot4+2\cdot4+\frac{1}{2}\cdot2\cdot4=16\,\text{m}\). Thus the maximum altitude is \(150\,\text{m}+16\,\text{m}=166\,\text{m}\). 4. The signed area from \(t=7\) to \(t=10\) is \(-\left(\frac{1}{2}\cdot2\cdot2+1\cdot2\right)=-4\,\text{m}\). Therefore, the final altitude is \(166\,\text{m}-4\,\text{m}=162\,\text{m}\).

a) From \(t=6\) to \(t=7\) minutes b) From \(t=6\) to \(t=7\) minutes; maximum altitude \(166\,\text{m}\) c) \(162\,\text{m}\)

- Areas above and below the time axis have different signs when finding displacement. - For total distance, use the magnitude of every geometric area. - Break the graph into triangles and a rectangle.

1. The signed area above the time axis is a triangle, a rectangle, and a triangle: \(\frac{1}{2}\cdot2\cdot4+4\cdot4+\frac{1}{2}\cdot2\cdot4=24\,\text{m}\). 2. The signed area below the time axis is \(-\frac{1}{2}\cdot2\cdot4=-4\,\text{m}\). 3. The displacement is the total signed area: \(24-4=20\,\text{m}\). 4. Total distance counts both regions positively: \(24+4=28\,\text{m}\). 5. The results differ because displacement includes direction, while total distance does not.

a) \(20\,\text{m}\) b) \(28\,\text{m}\) c) Displacement uses signed area, while total distance uses the magnitudes of all areas.

- How is height reconstructed from vertical velocity? - Find an antiderivative of \(v(t)\) to describe the change in height. - Include the initial height of \(150\,\text{m}\). - When does velocity change from negative to positive? - To find a return to the initial height, set the accumulated change in height equal to zero.

1. The height function is \(h(t)=150+\int_0^t v(x)\,\text{d}x=150-0.05t^4+1.4t^3-8t^2\). 2. Evaluating gives \(h(4)=98.8\,\text{m}\) and \(h(10)=250\,\text{m}\). 3. The velocity changes from negative to positive at \(t=5\), so the height has a minimum there. \(h(5)=93.75\,\text{m}\). 4. Setting \(h(t)=150\) gives \(-0.05t^2(t^2-28t+160)=0\). The positive solutions are \(t=8\) and \(t=20\). On the modeled interval, the first return occurs at \(t=8\) minutes.

a) \(h(4)=98.8\,\text{m}\) and \(h(10)=250\,\text{m}\) b) At \(t=5\) minutes; the minimum height is \(93.75\,\text{m}\). c) \(8\) minutes

- How are position, velocity, and time related? - What does an intersection of the velocity graphs mean? - How do you find distance from a velocity graph or function? - How can the difference of two accumulated distances be written as one integral?

1. Set the velocities equal: \(-\frac{1}{4}t^2+2t+2=\frac{1}{2}t+2\). This simplifies to \(-\frac{1}{4}t^2+\frac{3}{2}t=0\), so the positive solution is \(t=6\). The common velocity is \(5\,\text{m/s}\). 2. Paul travels \(\int_0^6\left(-\frac{1}{4}t^2+2t+2\right)\,\text{d}t=[-\frac{1}{12}t^3+t^2+2t]_0^6=30\,\text{m}\). 3. Jordan travels \(\int_0^6\left(\frac{1}{2}t+2\right)\,\text{d}t=[\frac{1}{4}t^2+2t]_0^6=21\,\text{m}\). 4. Paul is ahead by \(30-21=9\,\text{m}\). This lead equals \(\int_0^6(v_p(t)-v_j(t))\,\text{d}t\), the area between the graphs on \([0,6]\).

a) At \(t=6\,\text{s}\); the common velocity is \(5\,\text{m/s}\). b) Paul: \(30\,\text{m}\); Jordan: \(21\,\text{m}\) c) Paul is ahead by \(9\,\text{m}\), equal to the area between the velocity graphs on \([0,6]\).

- First determine the equation of the line in the velocity graph. - How are velocity and position related? - Use the known position at \(t=6\) to determine the integration constant.

1. The velocity graph is a line with y-intercept \(1\) and slope \(\frac{2-1}{2-0}=\frac{1}{2}\). Thus, \(v(t)=\frac{1}{2}t+1\). 2. Integrate velocity to obtain position: \(s(t)=\frac{1}{4}t^2+t+C\). 3. Use \(s(6)=30\): \(\frac{1}{4}(6)^2+6+C=30\). 4. Since \(9+6+C=30\), \(C=15\). 5. Therefore, \(s(t)=\frac{1}{4}t^2+t+15\).

\(s(t)=\frac{1}{4}t^2+t+15\)

- What does negative velocity indicate about direction of motion? - How is distance traveled found from a velocity-versus-time graph? - Distinguish total distance from net displacement. - Divide the region into triangles and rectangles.

1. The car moves away from the starting point while \(v(t)>0\), through \(t=12\). After that, velocity is negative, so the car moves back toward the start. Thus, the maximum distance occurs at \(t=12\). 2. The positive area from \(t=0\) to \(t=12\) is \(\frac{1}{2}(4)(10)+(4)(10)+\frac{1}{2}(4)(10)=80\,\text{m}\). 3. From \(t=12\) to \(t=20\), the signed area is \(-\frac{1}{2}(8)(5)=-20\,\text{m}\). 4. Total distance is the sum of the absolute areas: \(80+20=100\,\text{m}\). 5. The final position is the net displacement: \(80-20=60\,\text{m}\) from the starting point in the positive direction.

a) At \(t=12\,\text{s}\); the maximum distance is \(80\,\text{m}\). b) \(100\,\text{m}\) c) \(60\,\text{m}\) from the starting point in the positive direction

- What physical quantity is represented by the integral of velocity over time? - How does the sign of velocity affect the signed area? - Use formulas for the areas of a trapezoid and a triangle. - Identify when the vehicle changes direction.

1. The definite integral of velocity gives the vehicle's net change in position over the interval. 2. The area above the time axis from \(t=0\) to \(t=8\) is a trapezoid with area \(\frac{1}{2}(8+4)(10)=60\,\text{m}\). 3. The area below the time axis from \(t=8\) to \(t=10\) is a triangle with magnitude \(\frac{1}{2}(2)(10)=10\,\text{m}\). 4. Therefore, \(\int_0^{10}v(t)\,\text{d}t=60-10=50\,\text{m}\). 5. The vehicle moves away from its starting point while \(v(t)>0\), through \(t=8\). After \(t=8\), the velocity is negative, so it moves back toward the start. Thus, it is farthest away at \(t=8\).

a) The integral is the vehicle's net change in position. b) \(50\,\text{m}\) c) \(t=8\,\text{s}\)

- Because the velocity is nonnegative, integrate velocity to obtain distance. - Keep the exponential expression exact until the requested rounding step. - To solve for the threshold time, isolate the exponential term before taking a logarithm. - For the limiting velocity, examine what happens to the exponential term for large \(t\).

1. Since \(v(t)\ge0\), distance traveled on \([0,10]\) is \(\int_0^{10}40(1-e^{-0.25t})\,\text{d}t\). 2. An antiderivative is \(40t+160e^{-0.25t}\), so the distance is \(240+160e^{-2.5}\approx253.13\,\text{m}\). 3. Solve \(20=40(1-e^{-0.25t})\). Then \(e^{-0.25t}=\frac{1}{2}\), so \(t=4\ln2\approx2.77\,\text{s}\). 4. As \(t\to\infty\), \(e^{-0.25t}\to0\), so \(v(t)\to40\,\text{m/s}\).

a) \(240+160e^{-2.5}\,\text{m}\approx253.13\,\text{m}\) b) \(4\ln2\,\text{s}\approx2.77\,\text{s}\) c) \(40\,\text{m/s}\)

- When is a moving object's velocity zero? - How are velocity and position related? - Use the sign of velocity to determine direction. - For total distance, add the absolute changes in position over intervals with a constant direction.

1. Solve \(v(t)=0\): \(\frac{1}{2}t^2-4t+6=0\), or \(t^2-8t+12=0\). Thus, \(t=2\) and \(t=6\). Velocity is positive on \([0,2)\), negative on \((2,6)\), and positive on \((6,8]\). 2. A position function satisfying \(s(0)=0\) is \(s(t)=\frac{1}{6}t^3-2t^2+6t\). Therefore, \(s(6)=0\), so the vehicle is back at the starting point. 3. \(s(2)=\frac{16}{3}\,\text{m}\approx5.33\,\text{m}\), so that is its distance from the start at \(t=2\). 4. The total distance is \(|s(2)-s(0)|+|s(6)-s(2)|+|s(8)-s(6)|=\frac{16}{3}+\frac{16}{3}+\frac{16}{3}=16\,\text{m}\).

a) The vehicle is at rest at \(t=2\,\text{s}\) and \(t=6\,\text{s}\). It moves in the positive direction on \([0,2)\), in the negative direction on \((2,6)\), and in the positive direction on \((6,8]\). b) \(s(6)=0\,\text{m}\); it is at the starting point. c) \(\frac{16}{3}\,\text{m}\approx5.33\,\text{m}\) d) \(16\,\text{m}\)

- How does area under a velocity-versus-time graph relate to distance? - Divide the region into triangles and a rectangle. - Average speed is total distance divided by total time. - Convert meters per second to miles per hour after finding the average. - Determine how much distance is covered in each phase.

1. From \(0\) to \(10\) seconds, the vehicle accelerates at a constant rate. From \(10\) to \(50\) seconds, it travels at a constant \(20\,\text{m/s}\). From \(50\) to \(60\) seconds, it decelerates at a constant rate to rest. 2. The total distance is the area under the graph: \(\frac{1}{2}(10)(20)+(40)(20)+\frac{1}{2}(10)(20)=1000\,\text{m}\). 3. The average speed is \(\frac{1000}{60}=\frac{50}{3}\,\text{m/s}\). Using \(1\,\text{m/s}\approx2.23694\,\text{mph}\), this is approximately \(37.3\,\text{mph}\). 4. Half the distance is \(500\,\text{m}\). The first phase covers \(100\,\text{m}\), leaving \(400\,\text{m}\) at \(20\,\text{m/s}\). That takes \(20\) more seconds, so half the distance is reached at \(t=30\,\text{s}\).

a) \(0\) to \(10\,\text{s}\): constant acceleration; \(10\) to \(50\,\text{s}\): constant velocity; \(50\) to \(60\,\text{s}\): constant deceleration to rest b) \(1000\,\text{m}\) c) Approximately \(37.3\,\text{mph}\) d) \(t=30\,\text{s}\)

- Read the line's vertical intercept and slope from the graph. - For the geometric approach, identify the two parallel sides of the trapezoid under the line from \(0\) to \(x\). - For the calculus check, integrate the velocity formula you obtained from the graph. - Before calling displacement “distance,” check the sign of the velocity over the interval.

1. The graph is a line with \(v(0)=2\) and slope \(0.4\), so \(v(x)=0.4x+2\). 2. The region under the velocity graph from \(0\) to \(x\) is a trapezoid with parallel sides \(2\) and \(0.4x+2\), separated by \(x\). Thus \(s(x)=\frac{2+(0.4x+2)}{2}x=0.2x^2+2x\). 3. Integrating the velocity formula gives \(\int_0^x(0.4t+2)\,\text{d}t=0.2x^2+2x\), confirming the geometric result. 4. The velocity is positive on \([0,10]\), so distance equals displacement. Therefore, \(s(10)=0.2(10)^2+2(10)=40\,\text{ft}\).

a) \(s(x)=0.2x^2+2x\) b) \(v(t)=0.4t+2\), and integration gives \(s(x)=0.2x^2+2x\) c) \(40\,\text{ft}\)

- Acceleration accumulates into a change in velocity. - Use each actual time width in the trapezoidal estimate.

1. On \([0,2]\), the trapezoidal estimate is \(\frac{2}{2}(1+3)=4\,\text{m/s}\). 2. On \([2,5]\), the trapezoidal estimate is \(\frac{3}{2}(3+2)=\frac{15}{2}\,\text{m/s}\). 3. On \([5,7]\), the trapezoidal estimate is \(\frac{2}{2}(2-1)=1\,\text{m/s}\). 4. The estimated change in velocity is \(4+\frac{15}{2}+1=\frac{25}{2}\,\text{m/s}\). 5. Add the initial velocity: \(v(7)\approx5+\frac{25}{2}=\frac{35}{2}\,\text{m/s}\).

\(v(7)\approx \frac{35}{2}\,\text{m/s}\)

- Work backward from the later position. - The integral of velocity gives the signed position change.

1. The displacement is \(\int_0^4v(t)\,\text{d}t=\frac{8}{\pi} + 12\,\text{m}\). 2. Since \(s(4)=s(0)+\frac{8}{\pi} + 12\), solve \(25=s(0)+\frac{8}{\pi} + 12\). 3. Thus \(s(0)=13 - \frac{8}{\pi}\,\text{m}\).

\(13 - \frac{8}{\pi}\,\text{m}\)

- Build the position function from the velocity and initial position. - Passing through the origin is a condition on position, not velocity. - Check for a direction change between the two crossing times before finding distance.

1. Integrating velocity gives \(s(t)=t^2-4t+3\). 2. Solve \(s(t)=0\): \((t-1)(t-3)=0\), so the particle is at the origin at \(t=1\) and \(t=3\). 3. The velocity changes sign at \(t=2\), where \(s(2)=-1\). 4. From \(t=1\) to \(t=2\), the particle travels \(1\) unit, and from \(t=2\) to \(t=3\), it travels another \(1\) unit. 5. The total distance between the origin crossings is \(2\) units.

The particle passes through the origin at \(t=1\) and \(t=3\), and it travels \(2\) units between those times.

- Average velocity uses signed displacement. - Average speed uses total distance, so locate direction changes first.

1. The displacement is \(\int_0^8v(t)\,\text{d}t=\frac{32}{3}\,\text{m}\), so the average velocity is \(\frac{4}{3}\,\text{m/s}\). 2. The sign changes occur at \(t=2\) and \(t=6\). 3. The total distance is \(\int_0^2v(t)\,\text{d}t-\int_2^6v(t)\,\text{d}t+\int_6^8v(t)\,\text{d}t=32\,\text{m}\). 4. The average speed is \(4\,\text{m/s}\).

Average velocity: \(\frac{4}{3}\,\text{m/s}\) Average speed: \(4\,\text{m/s}\)

- Locate where the velocity changes sign. - Treat traveled distance on each direction interval as positive.

1. The velocity changes sign at \(t=4\), so the absolute value cannot be applied only after combining the signed areas. 2. Distance on \([0,4]\) is \(\int_0^4(4-t)\,\text{d}t=8\). 3. Distance on \([4,7]\) is \(-\int_4^7(4-t)\,\text{d}t=\frac92\). 4. The total distance is \(8+\frac92=\frac{25}{2}\).

Wiremu found the magnitude of displacement, not total distance. The correct distance is \(\frac{25}{2}\).

- Compare corresponding pieces on the left and right halves of the graph. - Signed area determines displacement, while unsigned area determines distance. - Use trapezoid areas rather than finding formulas for the line segments.

1. The velocity graph on \([4,8]\) is the negative of the graph on \([0,4]\) reflected across \(t=4\). 2. The positive signed area is \(\frac12(2+3)(2)+\frac12(3+0)(2)=5+3=8\). 3. The signed area on \([4,8]\) is \(-8\), so the displacement on \([0,8]\) is \(0\). Thus \(s(8)=s(0)\). 4. The total distance is \(8+8=16\) units.

\(s(8)=s(0)\), and the total distance traveled is \(16\) units.

- Use the sign of velocity to decide when position is decreasing or increasing. - Look for the point where the direction of motion changes. - Relate the signed area under the velocity graph to the change in position.

1. Position decreases while \(v(t)<0\) and increases while \(v(t)>0\). 2. The graph crosses \(v=0\) at \(t=3\), changing from negative to positive, so the minimum position occurs at \(t=3\). 3. From \(t=0\) to \(t=2\), the signed area is \(-4\,\text{m}\). From \(t=2\) to \(t=3\), the triangular signed area is \(-1\,\text{m}\). 4. Thus \(s(3)=6-4-1=1\,\text{m}\).

The position is smallest at \(t=3\,\text{s}\), and the minimum position is \(1\,\text{m}\).

- How are vertical velocity and height related? - How does the initial height enter the accumulation function? - At what times does the balloon change vertical direction? - To find an absolute maximum, compare critical points and endpoints. - What equation expresses a return to the initial height?

1. The height is \(h(t)=250+\int_0^t v(x)\,\text{d}x=250+\frac{t^4}{200{,}000}-\frac{13t^3}{15{,}000}+0.03t^2\). 2. Evaluating gives \(h(60)=235.6\,\text{m}\) and \(h(120)=221.2\,\text{m}\). 3. Since \(h'(t)=v(t)\), the interior critical times are \(t=30\) and \(t=100\). Compare the heights at \(t=0\), \(30\), \(100\), and the endpoint \(150\). The values are \(250\), \(257.65\), approximately \(183.33\), and \(531.25\) meters, respectively. Therefore, the greatest height on \([0,150]\) is \(531.25\,\text{m}\) at \(t=150\). 4. Substitution gives \(h(40)=255.333\ldots\,\text{m}\approx255.3\,\text{m}\). 5. Setting \(h(t)=250\) gives \(t^2\left(\frac{t^2}{4}-\frac{130t}{3}+1500\right)=0\). The positive solutions are \(t=\frac{260-20\sqrt{34}}{3}\approx47.8\) and \(t=\frac{260+20\sqrt{34}}{3}\approx125.5\). The first return occurs at approximately \(47.8\,\text{s}\).

a) After \(1\) minute: \(235.6\,\text{m}\); after \(2\) minutes: \(221.2\,\text{m}\) b) At \(t=150\,\text{s}\); the greatest height is \(531.25\,\text{m}\). c) \(h(40)\approx255.3\,\text{m}\) d) Approximately \(47.8\,\text{s}\) after launch

- How are velocity and acceleration related? - Which differentiation rules apply to a product of a polynomial and an exponential function? - What physical quantity results from integrating velocity over time? - To find an antiderivative of a polynomial times an exponential function, consider integration by parts or an undetermined-coefficients form.

1. Acceleration is the derivative of velocity. Using the product and chain rules gives \(a(t)=v'(t)=(t^2-14t+20)e^{-0.5t}\). 2. \(a(2)=-4e^{-1}\approx-1.47\,\text{m/s}^2\). 3. The depth after \(10\) seconds is \(s(10)=\int_0^{10}(20t-2t^2)e^{-0.5t}\,\text{d}t\). 4. An antiderivative is \(V(t)=(4t^2-24t-48)e^{-0.5t}\), so \(s(10)=[(4t^2-24t-48)e^{-0.5t}]_0^{10}=112e^{-5}+48\approx48.75\,\text{m}\). 5. Because \(v(t)\ge0\) on \([0,10]\), the integral is both the downward displacement and the total vertical distance traveled during the interval.

a) Approximately \(-1.47\,\text{m/s}^2\) b) Approximately \(48.75\,\text{m}\) c) It is the vehicle's downward displacement, which is also its depth after \(10\) seconds because the velocity is nonnegative throughout the interval.

- When is a fraction with a fixed positive numerator largest? - Distance traveled is the area under a velocity-versus-time graph. - The area under a line can be found geometrically or by integration. - When does the distance between two runners increase, and when does it decrease?

1. Since \((t-5)^2\ge0\), the denominator \((t-5)^2+1\) is smallest at \(t=5\). Therefore, the fraction is largest there, and \(f(5)=2+6=8\,\text{m/s}\). 2. Ben travels \(\int_0^{12}(0.4t+1)\,\text{d}t=[0.2t^2+t]_0^{12}=40.8\,\text{m}\). 3. Alex travels \(\int_0^{12}\left(2+\frac{6}{(t-5)^2+1}\right)\,\text{d}t=[2t+6\arctan(t-5)]_0^{12}\approx40.814\,\text{m}\). 4. Alex is ahead by approximately \(40.814-40.8=0.014\,\text{m}\) at \(t=12\). 5. Alex's lead increases while \(f(t)>g(t)\) and decreases after \(g(t)>f(t)\). Thus, the greatest lead occurs where the velocity graphs intersect and switch order, at approximately \(t=6.62\,\text{s}\).

a) The denominator is smallest at \(t=5\), so the maximum velocity is \(f(5)=8\,\text{m/s}\). b) Alex travels approximately \(40.814\,\text{m}\), and Ben travels \(40.8\,\text{m}\). Alex is ahead by approximately \(0.014\,\text{m}\). c) The maximum lead occurs where the velocity graphs intersect and Alex changes from being faster to being slower, at approximately \(t=6.62\,\text{s}\).

- Which operation relates velocity to distance traveled? - Use the first derivative to locate a maximum of \(v(t)\). - Use integration by parts for a linear factor times an exponential function. - For the total distance, determine the limiting value of the antiderivative term as \(t\to\infty\).

1. Using the product and chain rules, \(v'(t)=(1-0.5t)e^{-0.5t}\). The only positive critical time is \(t=2\), and the derivative changes from positive to negative there. Thus, the maximum velocity is \(v(2)=2e^{-1}\approx0.74\,\text{m/s}\). 2. An antiderivative is \((-2t-4)e^{-0.5t}\). Therefore, \(\int_0^6te^{-0.5t}\,\text{d}t=[(-2t-4)e^{-0.5t}]_0^6=4-16e^{-3}\approx3.20\,\text{m}\). 3. The total distance is \(\int_0^\infty te^{-0.5t}\,\text{d}t=\lim_{b\to\infty}[(-2t-4)e^{-0.5t}]_0^b=4\,\text{m}\), because \((-2b-4)e^{-0.5b}\to0\).

a) At \(t=2\,\text{s}\); the maximum velocity is \(2e^{-1}\,\text{m/s}\approx0.74\,\text{m/s}\). b) \(4-16e^{-3}\,\text{m}\approx3.20\,\text{m}\) c) \(4\,\text{m}\)

- Accumulated acceleration changes velocity. - Break the graph into signed geometric areas. - A maximum velocity occurs when the way velocity is changing switches from increasing to decreasing.

1. The signed area under \(a\) from \(0\) to \(6\) is \(4+2-2=4\,\text{m/s}\). 2. Therefore, \(v(6)=2+4=6\,\text{m/s}\). 3. Velocity increases while \(a(t)>0\) and decreases while \(a(t)<0\). The graph crosses \(a=0\) at \(t=\frac72\). 4. The accumulated acceleration from \(0\) to \(\frac72\) is \(4+\frac12\left(\frac32\right)(3)=\frac{25}{4}\,\text{m/s}\). 5. The greatest velocity is \(2+\frac{25}{4}=\frac{33}{4}\,\text{m/s}\) at \(t=\frac72\,\text{s}\).

a) \(v(6)=6\,\text{m/s}\) b) The maximum occurs at \(t=\frac72\,\text{s}\), and the maximum velocity is \(\frac{33}{4}\,\text{m/s}\).

- Focus on how the difference between the two positions changes. - The difference of the velocities gives the rate of change of separation. - After finding a separation function, check whether its sign affects the distance.

1. The signed separation is \(D(t)=s_A(t)-s_B(t)=\int_0^t[v_A(u)-v_B(u)]\,\text{d}u\). 2. Since \(v_A-v_B=6-2t\), \(D(t)=6t-t^2\). 3. On \([0,5]\), \(D(t)\ge0\), so the distance apart is \(D(t)\). 4. The quadratic reaches its maximum at \(t=3\). 5. The maximum separation is \(D(3)=9\) units.

They are farthest apart at \(t=3\), with maximum separation \(9\) units.

- Use the velocity information to determine the constant after accumulating acceleration. - The known position is at an interior time, so accumulate forward and backward from that time. - Keep the position and velocity constants separate.

1. Integrating acceleration gives \(v(t)=\frac32t^2-6t+C\). 2. The condition \(v(2)=-1\) gives \(C=5\), so \(v(t)=\frac32t^2-6t+5\). 3. An antiderivative of velocity is \(F(t)=\frac12t^3-3t^2+5t\), with \(F(2)=2\). 4. Since \(s(t)=5+F(t)-F(2)=F(t)+3\), \(s(0)=3\) and \(s(4)=7\).

\(s(0)=3\) and \(s(4)=7\)

- Translate “returns to the origin” into a condition on net displacement. - After finding the parameter, locate the velocity sign change before computing distance.

1. Returning to the origin requires zero displacement: \(\int_0^6(k-t)\,\text{d}t=6k-18=0\), so \(k=3\). 2. With \(v(t)=3-t\), the direction changes at \(t=3\). 3. The distance is \(\int_0^3(3-t)\,\text{d}t-\int_3^6(3-t)\,\text{d}t=\frac92+\frac92=9\).

\(k=3\), and the total distance is \(9\) units.

- Represent velocity as accumulated constant acceleration. - Use both the final velocity and the integrated position change.

1. Let the constant acceleration be \(a\) and initial velocity be \(v_0\). Then \(v(t)=v_0+at\). 2. The final velocity gives \(v_0+5a=14\). 3. The position change gives \(50=\int_0^5(v_0+at)\,\text{d}t=5v_0+\frac{25}{2}a\). 4. Solving the system gives \(v_0=6\,\text{m/s}\) and \(a=\frac{8}{5}\,\text{m/s}^2\).

Initial velocity: \(6\,\text{m/s}\) Acceleration: \(\frac{8}{5}\,\text{m/s}^2\)