# College Physics 2.24 – Motion of a car entering a freeway

## Solution:

### Part A

The sketch of the situation is shown below. Also, the knowns and unknowns are in the illustration.

From the illustration above, we can see that the initial velocity is 0 m/s, the initial time and initial distance are also zero. The final velocity and the final distance are unknowns. The time at the final location is 12 seconds and the acceleration is constant all throughout the trip at 2.40 meters per second square.

### Part B

The knowns are:

$\displaystyle \text{a}=2.40\:\text{m/s}^2,\:\text{t}=12.0\:\sec ,\:\text{v}_0=0\:\text{m/s},\:\text{x}_0=0\:\text{m}$

### Part C

For this part, the unknown is the value of x. If we examine the equations for constant acceleration and the given values in this problem, we can readily use the formula $\text{x}=\text{x}_0+\text{v}_0\text{t}+\frac{1}{2}\text{at} ^2$. That is

$\text{x}=\text{x}_0+\text{v}_0\text{t}+\frac{1}{2}\text{at} ^2$

$\displaystyle \text{x}=0\:\text{m}+\left(0\:\text{m/s}\right)\left(12.0\:\text{s}\right)+\frac{1}{2}\left(2.40\:\text{m/s}^2\right)\left(12.0\:\text{s}\right)^2$

$\displaystyle \text{x}=172.8\:\text{m}$

### Part D

For this part, the unknown is the value of v. The equation that can be used based from the known variables is $\displaystyle \text{v}=\text{v}_0+\text{at}$. That is

$\displaystyle \text{v}=\text{v}_0+\text{at}$

$\displaystyle \text{v}=0\:\text{m/s}+\left(2.40\:\text{m/s}^2\right)\left(12.0\:\text{s}\right)$

$\displaystyle \text{v}=28.8\:\text{m/s}$