College Physics 2.24 – Motion of a car entering a freeway

While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s.

(a) Draw a sketch of the situation.

(b) List the knowns in this problem.

(c) How far does the car travel in those 12.0 s? To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, check your units, and discuss whether the answer is reasonable.

(d) What is the car’s final velocity? Solve for this unknown in the same manner as in part (c), showing all steps explicitly.


Part A

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

College Physics Problem 2.24 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}