Chapter 2, Problem 13P

A particle starts from rest and accelerates as shown in Figure P2.13. Determine (a) the particle’s speed at t = 10.0 s and at t = 20.0 s, and (b) the distance traveled in the first 20.0 s.

To determine

The particle’s speed at $t=10.0\text{ s}$ and at $t=20.0\text{ s}$.

Answer to Problem 13P

The particle’s speed at $t=10.0\text{ s}$ is $20\text{ m/s}$ and at $t=20.0\text{ s}$ is $5\text{ m/s}$.

Explanation of Solution

Write the equation for the final velocity of a particle.

  $v=v_0+\Delta v$        (I)

Here, $v$ is the final velocity of a particle, $v_0$ is the initial velocity of a particle and $\Delta v$ is the change in velocity.

The area under acceleration versus time graph under specific time intervals gives the change in velocity of the particle during the time interval. Speed is the magnitude of velocity.

The sides of a unit rectangle in the graph are $1{\text{ m/s}}^2$ and $5\text{ s}$ so that the area of a unit rectangle will be $\left(1{\text{ m/s}}^2\right)\left(5\text{ s}\right)=5\text{ m/s}$ .

The area under the graph in the interval $0<t<10.0\text{ s}$ has four unit rectangles so that the area under the curve in the time interval is $4\left(5\text{ m/s}\right)=20\text{ m/s}$ .

Since the particle starts from rest, its velocity at $t=0\text{ s}$ will be $0\text{ m/s}$.

Substitute $0\text{ m/s}$ for $v_0$ and $20\text{ m/s}$ for $\Delta v$ in equation (I) to find $v$ at $t=10.0\text{ s}$.

  $\begin{array}{c}v=0\text{ m/s}+20\text{ m/s}\\ =20\text{ m/s}\end{array}$

The area under the graph in the interval $10.0<t<20.0\text{ s}$ has three unit rectangles in the negative direction so that the area under the curve in the time interval is $-3\left(5\text{ m/s}\right)=-15\text{ m/s}$.

Substitute $20\text{ m/s}$ for $v_0$ and $-15\text{ m/s}$ for $\Delta v$ in equation (I) to find $v$ at $t=20.0\text{ s}$.

  $\begin{array}{c}v=20\text{ m/s}+\left(-15\text{ m/s}\right)\\ =5\text{ m/s}\end{array}$

Conclusion:

Therefore, the particle’s speed at $t=10.0\text{ s}$ is $20\text{ m/s}$ and that at $t=20.0\text{ s}$ is $5\text{ m/s}$.

To determine

The distance travelled in the first $20.0\text{ s}$.

Answer to Problem 13P

The distance travelled in the first $20.0\text{ s}$ is $263\text{ m}$.

Explanation of Solution

The area under velocity versus time graph under specific time intervals gives the displacements during the time interval.

In part (a), it is found that the velocity at $t=0\text{ s}$ is $0\text{ m/s}$, at $t=10.0\text{ s}$ is $20\text{ m/s}$ and that at $t=20.0\text{ s}$ is $5\text{ m/s}$ . In the given acceleration versus time graph, the acceleration is zero in the interval $10.0\text{ s}<t<15.0\text{ s}$ . This implies there is no velocity change during the interval or the velocity at $t=15.0\text{ s}$ is same as velocity at $t=10.0\text{ s}$ . These data can be used to plot the velocity versus time graph for the given time interval.

The velocity versus time graph is shown below.

The area from $0-10\text{ s}$ is that of a triangle.

Write the equation for the area of a triangle.

  $A_t=\frac{1}{2}bh$        (II)

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

In figure 1, the base of the triangle from $0-10\text{ s}$ is $10\text{ s}$ and the height of the triangle is $20\text{ m/s}$ .

Substitute $10\text{ s}$ for $b$ and $20\text{ m/s}$ for $h$ in equation (II) to find $A_t$ .

  $\begin{array}{c}{A}_{t_{0-10}}=\frac{1}{2}\left(10\text{ s}\right)\left(20\text{ m/s}\right)\\ =100\text{ m}\end{array}$

Here, $A_{t_{0-10}}$ is the area under the graph from $0-10\text{ s}$ .

The area from $10-15\text{ s}$ is that of a rectangle.

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.

In figure 1, the length of the rectangle from $10-15\text{ s}$ is $5\text{ s}$ and the breadth of the rectangle is $20\text{ m/s}$ .

Substitute $5\text{ s}$ for $l$ and $20\text{ m/s}$ for $d$ in equation (III) to find $A_r$ .

  $\begin{array}{c}{A}_{r_{10-15}}=\left(5\text{ s}\right)\left(20\text{ m/s}\right)\\ =100\text{ m}\end{array}$

Here, $A_{r_{10-15}}$ is the area under the graph from $10-15\text{ s}$ .

The area from $15-20\text{ s}$ is the sum of the area of a triangle and a rectangle.

In figure 1, the base of the triangle from $15-20\text{ s}$ is $5\text{ s}$ and the height of the triangle is $20\text{ m/s}-5\text{ m/s}=15\text{ m/s}$ .

Substitute $5\text{ s}$ for $b$ and $15\text{ m/s}$ for $h$ in equation (II) to find $A_t$ .

  $\begin{array}{c}{A}_{t_{15-20}}=\frac{1}{2}\left(5\text{ s}\right)\left(15\text{ m/s}\right)\\ =37.5\text{ m}\end{array}$

Here, $A_{t_{15-20}}$ is the area of the triangle under the graph from $15-20\text{ s}$ .

In figure 1, the length of the rectangle from $15-20\text{ s}$ is $5\text{ s}$ and the breadth of the rectangle is $5\text{ m/s}$ .

Substitute $5\text{ s}$ for $l$ and $5\text{ m/s}$ for $d$ in equation (III) to find $A_r$ .

  $\begin{array}{c}{A}_{r_{15-20}}=\left(5\text{ s}\right)\left(5\text{ m/s}\right)\\ =25\text{ m}\end{array}$

Here, $A_{r_{15-20}}$ is the area of the rectangle under the graph from $15-20\text{ s}$ .

Write the equation for total distance travelled.

  $\Delta x=A_{t_{0-10}}+A_{r_{10-15}}+A_{t_{15-20}}+A_{r_{15-20}}$        (IV)

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

Conclusion:

Substitute $100\text{ m}$ for $A_{t_{0-10}}$, $100\text{ m}$ for $A_{r_{10-15}}$, $37.5\text{ m}$ for $A_{t_{15-20}}$ and $25\text{ m}$ for $A_{r_{15-20}}$ in equation (IV) to find $\Delta x$ .

  $\begin{array}{c}\Delta x=100\text{ m}+100\text{ m}+37.5\text{ m}+25\text{ m}\\ =263\text{ m}\end{array}$

Therefore, the distance travelled in the first $20.0\text{ s}$ is $263\text{ m}$.