Chapter 2, Problem 4SP

(a)

To determine

The time taken for the car to reach the velocity of $27\text{m/s}$.

Answer to Problem 4SP

The time taken for the car to reach the velocity of $27\text{m/s}$ is $8\text{s}$.

Explanation of Solution

Given info: Initial velocity of the car is $11\text{ m/s}$, acceleration is $2.0{\text{m/s}}^2$ and final velocity is $27\text{m/s}$.

Write the expression for Newton’s equation of motion.

$v=v_0+at$

Here,

v is the final velocity

$v_0$ is the initial velocity

t is the initial time

a is the acceleration

Re-write the above equation to get t.

$t=\frac{v-v_0}{a}$

Substitute $27\text{m/s}$ for v, $11\text{ m/s}$ for $v_0$ and $2.0{\text{m/s}}^2$ for a to get t.

$\begin{array}{c}t=\frac{\left(27\text{m/s}\right)-\left(11\text{m/s}\right)}{2.0{\text{m/s}}^{\text{2}}}\\ =8\text{s}\end{array}$

Conclusion:

Therefore, the time taken for the car to reach the velocity of $27\text{m/s}$ is $8\text{s}$.

(b)

To determine

The distance covered by the car.

Answer to Problem 4SP

The distance covered by the car is $152\text{m}$.

Explanation of Solution

Given info: Initial velocity of the car is $11\text{ m/s}$, acceleration is $2.0{\text{m/s}}^2$ and final velocity is $27\text{m/s}$.

Write the expression for distance travelled.

$s=v_{0}t+\frac{1}{2}a{t}^2$

Here,

$s$ is the distance

Substitute $8\text{s}$ for t, $11\text{ m/s}$ for $v_0$ and $2.0{\text{m/s}}^{\text{2}}$ for a to get $s$.

$\begin{array}{c}s=\left(11\text{m/s}\right)\left(8\text{s}\right)+\frac{1}{2}\left(2.0{\text{m/s}}^{\text{2}}\right){\left(8\text{s}\right)}^2\\ =152\text{m}\end{array}$

Conclusion:

Therefore, the distance covered by the car is $152\text{m}$.

(c)

To determine

The graph of distance versus time.

Answer to Problem 4SP

The distance is plotted against time in the graph.

Explanation of Solution

Write the expression for distance.

$s=v_{0}t+\frac{1}{2}a{t}^2$

At 0s,

Substitute 0s for t, $11\text{ m/s}$ for $v_0$ and $2.0{\text{m/s}}^{\text{2}}$ for a to get $s$.

$\begin{array}{c}s=\left(11\text{m/s}\right)\left(0\text{s}\right)+\frac{1}{2}\left(2.0{\text{m/s}}^{\text{2}}\right){\left(0\text{s}\right)}^2\\ =0\text{m}\end{array}$

At 1s,

Substitute 1s for t, $11\text{ m/s}$ for $v_0$ and $2.0{\text{m/s}}^{\text{2}}$ for a to get $s$.

$\begin{array}{c}s=\left(11\text{m/s}\right)\left(1\text{s}\right)+\frac{1}{2}\left(2.0{\text{m/s}}^{\text{2}}\right){\left(1\text{s}\right)}^2\\ =12\text{m}\end{array}$

At 2s,

Substitute 2s for t, $11\text{ m/s}$ for $v_0$ and $2.0{\text{m/s}}^{\text{2}}$ for a to get $s$.

$\begin{array}{c}s=\left(11\text{m/s}\right)\left(2\text{s}\right)+\frac{1}{2}\left(2.0{\text{m/s}}^{\text{2}}\right){\left(2\text{s}\right)}^2\\ =26\text{m}\end{array}$

At 3s,

Substitute 3s for t, $11\text{ m/s}$ for $v_0$ and $2.0{\text{m/s}}^{\text{2}}$ for a to get $s$.

$\begin{array}{c}s=\left(11\text{m/s}\right)\left(3\text{s}\right)+\frac{1}{2}\left(2.0{\text{m/s}}^{\text{2}}\right){\left(3\text{s}\right)}^2\\ =42\text{m}\end{array}$

At 4s,

Substitute 4s for t, $11\text{ m/s}$ for $v_0$ and $2.0{\text{m/s}}^{\text{2}}$ for a to get $s$.

$\begin{array}{c}s=\left(11\text{m/s}\right)\left(4\text{s}\right)+\frac{1}{2}\left(2.0{\text{m/s}}^{\text{2}}\right){\left(4\text{s}\right)}^2\\ =60\text{m}\end{array}$

At 5s,

Substitute 5s for t, $11\text{ m/s}$ for $v_0$ and $2.0{\text{m/s}}^{\text{2}}$ for a to get $s$.

$\begin{array}{c}s=\left(11\text{m/s}\right)\left(5\text{s}\right)+\frac{1}{2}\left(2.0{\text{m/s}}^{\text{2}}\right){\left(5\text{s}\right)}^2\\ =80\text{m}\end{array}$

At 6s,

Substitute 6s for t, $11\text{ m/s}$ for $v_0$ and $2.0{\text{m/s}}^{\text{2}}$ for a to get $s$.

$\begin{array}{c}s=\left(11\text{m/s}\right)\left(6\text{s}\right)+\frac{1}{2}\left(2.0{\text{m/s}}^{\text{2}}\right){\left(6\text{s}\right)}^2\\ =102\text{m}\end{array}$

At 7s,

Substitute 7s for t, $11\text{ m/s}$ for $v_0$ and $2.0{\text{m/s}}^{\text{2}}$ for a to get $s$.

$\begin{array}{c}s=\left(11\text{m/s}\right)\left(7\text{s}\right)+\frac{1}{2}\left(1.0{\text{m/s}}^{\text{2}}\right){\left(7\text{s}\right)}^2\\ =126\text{m}\end{array}$

At 8s,

Substitute 8s for t, $11\text{ m/s}$ for $v_0$ and $2.0{\text{m/s}}^{\text{2}}$ for a to get $s$.

$\begin{array}{c}s=\left(11\text{m/s}\right)\left(8\text{s}\right)+\frac{1}{2}\left(2.0{\text{m/s}}^{\text{2}}\right){\left(8\text{s}\right)}^2\\ =152\text{m}\end{array}$

The above values are plotted in the below graph.

Figure 1

Conclusion:

Distance is plotted against time in the graph.