Chapter 2, Problem 40P

An object is at x = 0 at t = 0 and moves along the x axis according to the velocity–time graph in Figure P2.40. (a) What is the object’s acceleration between 0 and 4.0 s? (b) What is the object’s acceleration between 4.0 s and 9.0 s? (c) What is the object’s acceleration between 13.0 s and 18.0 s? (d) At what time(s) is the object moving with the lowest speed? (e) At what time is the object farthest from x = 0? (f) What is the final position x of the object at t = 18.0 s? (g) Through what total distance has the object moved between t = 0 and t = 18.0 s?

Figure P2.40

To determine

The object’s acceleration between $0$ and $4.0\text{ s}$.

Answer to Problem 40P

The object’s acceleration between $0$ and $4.0\text{ s}$ is $0$.

Explanation of Solution

Acceleration is a measure of how rapidly the velocity is changing. It is defined as the change in velocity per unit time. The slope of the velocity versus time graph in a particular time interval gives the acceleration during that interval.

In the given velocity versus time graph, the curve is a straight parallel to the time axis between $0$ and $4.0\text{ s}$. This implies the velocity of the object is constant during the interval. Since there is no change in velocity, the acceleration will be zero during the time interval.

Conclusion:

Therefore, the object’s acceleration between $0$ and $4.0\text{ s}$ is $0$.

To determine

The object’s acceleration between $4.0\text{ s}$ and $9.0\text{ s}$.

Answer to Problem 40P

The object’s acceleration between $4.0\text{ s}$ and $9.0\text{ s}$ is $6.0{\text{ m/s}}^2$.

Explanation of Solution

The slope of the graph in the given interval gives the acceleration of the object during it.

Write the equation for the acceleration of the object between $4.0\text{ s}$ and $9.0\text{ s}$.

  $a=\frac{v_9-v_4}{9\text{ s}-4\text{ s}}$        (I)

Here, $a$ is the acceleration, $v_9$ is the velocity of the object at $9.0\text{ s}$ and $v_4$ is the velocity of the object at $4.0\text{ s}$.

Conclusion:

From the graph, the value of $v_9$ is $18\text{ m/s}$ and the value of $v_4$ is $12\text{ m/s}$.

Substitute $18\text{ m/s}$ for $v_9$ and $12\text{ m/s}$ for $v_4$ in equation (I) to find $a$.

  $\begin{array}{c}a=\frac{18\text{ m/s}-12\text{ m/s}}{9\text{ s}-4\text{ s}}\\ =6.0{\text{ m/s}}^2\end{array}$

Therefore, the object’s acceleration between $4.0\text{ s}$ and $9.0\text{ s}$ is $6.0{\text{ m/s}}^2$.

To determine

The object’s acceleration between $13.0\text{ s}$ and $18.0\text{ s}$.

Answer to Problem 40P

The object’s acceleration between $13.0\text{ s}$ and $18.0\text{ s}$ is $-3.6{\text{ m/s}}^2$.

Explanation of Solution

Write the equation for the acceleration of the object between $13.0\text{ s}$ and $18.0\text{ s}$.

  $a=\frac{v_{18}-v_{13}}{18\text{ s}-13\text{ s}}$        (II)

Here, $v_{18}$ is the velocity of the object at $18.0\text{ s}$ and $v_{13}$ is the velocity of the object at $13.0\text{ s}$.

Conclusion:

From the graph, the value of $v_{18}$ is $0\text{ m/s}$ and the value of $v_{13}$ is $18\text{ m/s}$.

Substitute $0\text{ m/s}$ for $v_{18}$ and $18\text{ m/s}$ for $v_{13}$ in equation (II) to find $a$.

  $\begin{array}{c}a=\frac{0\text{ m/s}-18\text{ m/s}}{18\text{ s}-13\text{ s}}\\ =-3.6{\text{ m/s}}^2\end{array}$

Therefore, the object’s acceleration between $13.0\text{ s}$ and $18.0\text{ s}$ is $-3.6{\text{ m/s}}^2$.

To determine

The times at which the object moves with the lowest speed.

Answer to Problem 40P

The times at which the object moves with the lowest speed are $6\text{ s}$ and $18\text{ s}$.

Explanation of Solution

In a velocity versus time graph, the velocity of an object will be plotted as a function of time. The velocity at a particular instant of time can be directly observed from the graph. Speed is the magnitude of velocity.

Speed can never be negative. The lowest possible value of speed is zero. In the graph, the velocity of the object is zero at $6\text{ s}$ and at $18\text{ s}$. This implies the speed of the object at these instants is zero.

Conclusion:

Therefore, the times at which the object moves with the lowest speed are $6\text{ s}$ and $18\text{ s}$.

To determine

The time at which the object is farthest from $x=0$.

Answer to Problem 40P

The time at which the object is farthest from $x=0$ is $18\text{ s}$.

Explanation of Solution

The given velocity versus time graph is of an object starting from $x=0$ and moving along the $x$ axis. The velocity of the object from $0\text{ s}$ to $6\text{ s}$ is negative. This implies during this time the object moves along the negative x direction into the negative coordinates. However since the magnitude of the velocity decreases after $4\text{ s}$ , it is clear that the object start to reverse its direction after $4\text{ s}$ . The object finally comes to $x=0$ at $t=6\text{ s}$ .

From $6\text{ s}$ onwards the velocity of the object increases to positive value. This implies the object is moving along positive $x$ direction. The object continue to move along the $+x$ direction up to $t=18\text{ s}$ . Since the particle continue to move along the $+x$ direction, the farthest distance from $x=0$ will be at the end of the object’s motion. It occurs at $t=18\text{ s}$ .

Conclusion:

Therefore, the time at which the object is farthest from $x=0$ is $18\text{ s}$.

To determine

The final position of the object at $t=18.0\text{ s}$.

Answer to Problem 40P

The final position of the object at $t=18.0\text{ s}$ is $84\text{ m}$.

Explanation of Solution

The cumulative area under the graph gives the maximum distance attained by the object.

The velocity versus time graph is shown below.

Write the equation for the area of a rectangle.

  $A_r=ld$        (III)

Here, $A_r$ is the area of the rectangle. $l$ is the length of the rectangle and $d$ is the breadth of the rectangle.

Write the equation for the area of a triangle.

  $A_t=\frac{1}{2}bh$        (IV)

Here, $A_t$ is the area of the triangle. $b$ is the base and $h$ is the height of the triangle.

The area from $0-4\text{ s}$ is that of a rectangle.

In figure 1, the length of the rectangle from $0-4\text{ s}$ is $4\text{ s}$ and the breadth of the rectangle is $-10\text{ m/s}$.

Substitute $4\text{ s}$ for $l$ and $-12\text{ m/s}$ for $d$ in equation (III) to find $A_r$.

  $\begin{array}{c}{A}_{r_{0-4}}=\left(4\text{ s}\right)\left(-12\text{ m/s}\right)\\ =-48\text{ m}\end{array}$

Here, $A_{r_{0-4}}$ is the area under the graph from $0-4\text{ s}$ .

The area from $4-6\text{ s}$ is that of a triangle.

In figure 1, the base of the triangle from $4-6\text{ s}$ is $2\text{ s}$ and the height of the triangle is $-12\text{ m/s}$ .

Substitute $2\text{ s}$ for $b$ and $-12\text{ m/s}$ for $h$ in equation (IV) to find $A_t$.

  $\begin{array}{c}{A}_{t_{4-6}}=\frac{1}{2}\left(2\text{ s}\right)\left(-12\text{ m/s}\right)\\ =-12\text{ m}\end{array}$

Here, $A_{t_{4-6}}$ is the area under the graph from $4-6\text{ s}$.

The area from $6-9\text{ s}$ is that of a triangle.

In figure 1, the base of the triangle from $6-9\text{ s}$ is $3\text{ s}$ and the height of the triangle is $18\text{ m/s}$.

Substitute $3\text{ s}$ for $b$ and $18\text{ m/s}$ for $h$ in equation (IV) to find $A_t$.

  $\begin{array}{c}{A}_{t_{6-9}}=\frac{1}{2}\left(3\text{ s}\right)\left(18\text{ m/s}\right)\\ =27\text{ m}\end{array}$

Here, $A_{t_{6-9}}$ is the area under the graph from $6-9\text{ s}$.

The area from $9-13\text{ s}$ is that of a rectangle.

In figure 1, the length of the rectangle from $9-13\text{ s}$ is $4\text{ s}$ and the breadth of the rectangle is $18\text{ m/s}$.

Substitute $4\text{ s}$ for $l$ and $18\text{ m/s}$ for $d$ in equation (III) to find $A_r$.

  $\begin{array}{c}{A}_{r_{9-13}}=\left(4\text{ s}\right)\left(18\text{ m/s}\right)\\ =72\text{ m}\end{array}$

Here, $A_{r_{9-13}}$ is the area under the graph from $9-13\text{ s}$.

The area from $13-18\text{ s}$ is that of a triangle.

In figure 1, the base of the triangle from $13-18\text{ s}$ is $5\text{ s}$ and the height of the triangle is $18\text{ m/s}$.

Substitute $5\text{ s}$ for $b$ and $18\text{ m/s}$ for $h$ in equation (III) to find $A_t$ .

  $\begin{array}{c}{A}_{t_{13-18}}=\frac{1}{2}\left(5\text{ s}\right)\left(18\text{ m/s}\right)\\ =45\text{ m}\end{array}$

Here, $A_{t_{13-18}}$ is the area under the graph from $13-18\text{ s}$.

Write the equation for the farthest distance.

  $\Delta x=A_{r_{0-4}}+A_{t_{4-6}}+A_{t_{6-9}}+A_{r_{9-13}}+A_{t_{13-18}}$        (V)

Here, $\Delta x$ is the farthest distance.

Conclusion:

Substitute $-48\text{ m}$ for $A_{r_{0-4}}$ , $-12\text{ m}$ for $A_{t_{4-6}}$, $27\text{ m}$ for $A_{t_{6-9}}$ , $72\text{ m}$ for $A_{r_{9-13}}$ and $45\text{ m}$ for $A_{t_{13-18}}$ in equation (V) to find $\Delta x$.

  $\begin{array}{c}\Delta x=\left(-48\text{ m}\right)+\left(-12\text{ m}\right)+27\text{ m}+72\text{ m}+45\text{ m}\\ =-60\text{ m}+144\text{ m}\\ =84\text{ m}\end{array}$

Therefore, the final position of the object at $t=18.0\text{ s}$ is $84\text{ m}$.

To determine

The total distance the object has moved between $t=0$ and $t=18.0\text{ s}$.

Answer to Problem 40P

The total distance the object has moved between $t=0$ and $t=18.0\text{ s}$ is $204\text{ m}$.

Explanation of Solution

The total distance travelled can be calculated by counting all contributions computed in part (f) as positive.

Conclusion:

Substitute $48\text{ m}$ for $A_{r_{0-4}}$ , $12\text{ m}$ for $A_{t_{4-6}}$, $27\text{ m}$ for $A_{t_{6-9}}$ , $72\text{ m}$ for $A_{r_{9-13}}$ and $45\text{ m}$ for $A_{t_{13-18}}$ in equation (V) to find the total distance travelled.

  $\begin{array}{c}\Delta x=48\text{ m}+12\text{ m}+27\text{ m}+72\text{ m}+45\text{ m}\\ =204\text{ m}\end{array}$

Here, $\Delta x$ is the total distance travelled.

Therefore, the total distance the object has moved between $t=0$ and $t=18.0\text{ s}$ is $204\text{ m}$.