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 d what is the speed of the car when they are abreast? express your answer with the appropriate units.

Explanation:

Step1: Find time for truck to move 60m

The truck starts from rest (\(u = 0\)), acceleration \(a_{truck}=2.10\ m/s^2\), displacement \(s = 60.0\ m\). Use the equation \(s=ut+\frac{1}{2}a_{truck}t^2\). Since \(u = 0\), the equation simplifies to \(s=\frac{1}{2}a_{truck}t^2\). Solving for \(t\):
\(t=\sqrt{\frac{2s}{a_{truck}}}=\sqrt{\frac{2\times60.0}{2.10}}\approx\sqrt{57.14}\approx7.56\ s\)

Step2: Find car's speed at that time

The car also starts from rest (\(u = 0\)), acceleration \(a_{car}=3.40\ m/s^2\). Use the equation \(v = u+at\) (since \(u = 0\), \(v=a_{car}t\)). Substitute \(t\approx7.56\ s\) and \(a_{car}=3.40\ m/s^2\):
\(v = 3.40\times7.56\approx25.7\ m/s\)

Answer:

\(25.7\ m/s\) (or more precisely, if we calculate \(t=\sqrt{\frac{120}{2.1}}\approx\sqrt{57.1429}\approx7.559\ s\), then \(v = 3.4\times7.559\approx25.7\ m/s\))