Chapter 3.2, Problem 19PP

To determine

To Calculate: The time taken by the car.

Answer to Problem 19PP

  $9.05\text{s}$

Explanation of Solution

Given:

The car slows down from $2\text{2 m/s}$ to $3.0\text{ m/s}$ at a constant acceleration of $2.1{\text{m/s}}^{\text{2}}$ .

Formula used:

Final velocity with constant acceleration is given by,

  $v_{final}=v_{initial}+a\Delta t$

Where, $\Delta t$ is the time taken, v indicates velocity and a is the acceleration.

Calculation:

Using the given values of initial velocity, final velocity, and acceleration, the time taken will be,

  $\begin{array}{c}{v}_{final}=v_{initial}+a\Delta t\\ 3.0=22+\left(-2.1\right)\left(\Delta t\right)\end{array}$

Negative sign is used since it’s a deceleration.

Then,

  $\begin{array}{c}\Delta t=\frac{3.0-22}{-2.1}\\ \approx 9.05\text{s}\end{array}$

Conclusion:

Thus, the time taken by the car to reach a velocity of $3.0\text{ m/s}$ from a velocity $2\text{2 m/s}$ is $9.05\text{s}$ .