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. how much time does it take the car to overtake the truck? express your answer with the appropriate units.

Explanation:

Step1: Analyze the truck's motion

The truck starts from rest (\(u = 0\)) and has a constant acceleration \(a_{truck}=2.10\space m/s^{2}\). The distance moved by the truck when overtaken is \(s_{truck} = 60.0\space m\). We use the kinematic equation \(s=ut+\frac{1}{2}at^{2}\). Since \(u = 0\), the equation simplifies to \(s_{truck}=\frac{1}{2}a_{truck}t^{2}\).

Step2: Solve for time \(t\)

We can re - arrange the equation \(s_{truck}=\frac{1}{2}a_{truck}t^{2}\) for \(t\). First, multiply both sides by \(2\) to get \(2s_{truck}=a_{truck}t^{2}\). Then divide both sides by \(a_{truck}\): \(t^{2}=\frac{2s_{truck}}{a_{truck}}\). Finally, take the square root of both sides: \(t=\sqrt{\frac{2s_{truck}}{a_{truck}}}\)

Substitute \(s_{truck} = 60.0\space m\) and \(a_{truck}=2.10\space m/s^{2}\) into the formula:

\(t=\sqrt{\frac{2\times60.0}{2.10}}=\sqrt{\frac{120}{2.10}}\approx\sqrt{57.14}\approx7.56\space s\)

We can also verify using the car's motion. The car has an acceleration \(a_{car}=3.40\space m/s^{2}\) and starts from rest. The distance moved by the car \(s_{car}=\frac{1}{2}a_{car}t^{2}\), and the distance between the car and the truck initially is \(d=s_{car}-s_{truck}\). But since we are interested in the time when the car overtakes the truck, the time taken for the truck to move \(60\space m\) under its acceleration is the same as the time taken for the car to cover the initial distance plus \(60\space m\). But from the truck's motion, we can directly calculate the time as above.

Answer:

The time taken is approximately \(\boldsymbol{7.56\space s}\) (the unit is seconds).