Question

a car and a truck start from rest at the same instant, with the car initially at some distance behind the truck. the truck has a constant acceleration of 2.10 m/s² and the car an acceleration of 3.40 m/s². the automobile overtakes the truck after the truck has moved 60.0 m. part b how far was the car behind the truck initially? express your answer with the appropriate units.

Explanation:

Step1: Find time for truck's motion

The truck starts from rest (\(u = 0\)) and moves a distance \(s_{truck}=60.0\space m\) with acceleration \(a_{truck}=2.10\space m/s^2\). Using the equation \(s = ut+\frac{1}{2}at^2\), since \(u = 0\), we have \(s_{truck}=\frac{1}{2}a_{truck}t^2\). Solving for \(t\):
\[
t=\sqrt{\frac{2s_{truck}}{a_{truck}}}=\sqrt{\frac{2\times60.0}{2.10}}\approx\sqrt{57.14}\approx7.56\space s
\]

Step2: Find distance traveled by car

The car also starts from rest (\(u = 0\)) with acceleration \(a_{car}=3.40\space m/s^2\) in time \(t\). Using \(s = ut+\frac{1}{2}at^2\) (again \(u = 0\)):
\[
s_{car}=\frac{1}{2}a_{car}t^2=\frac{1}{2}\times3.40\times(7.56)^2
\]
First, calculate \((7.56)^2\approx57.15\), then:
\[
s_{car}=\frac{1}{2}\times3.40\times57.15\approx1.70\times57.15\approx97.16\space m
\]

Step3: Find initial distance between car and truck

The initial distance \(d\) between the car and the truck is the difference between the distance the car travels and the distance the truck travels (since the car overtakes the truck after the truck has moved \(60.0\space m\)):
\[
d=s_{car}-s_{truck}=97.16 - 60.0 = 37.16\space m\approx37.2\space m
\]

Answer:

\(37.2\space m\)