Answer to Question #239599 in Physics for Adawiyah

Question

A scooter traveling on the straight road with speed of 30 km/h. Behind the scooter is a car travelling at a speed of 45 km/h. When the distance between them is 7.5km, the car is given acceleration of 15 km/h. Calculate the distance (of the scooter) and time taken when the car catches scooter.

Answer 1

Let's write the distances for scooter and car:

$$d_{sc}=v_{0,sc}t,$$ $$d_{car}+7.5=v_{0, car}t+\dfrac{1}{2}a_{car}t.$$

The car catches scooter when $d_{sc}=d_{car}$:

$$v_{0,sc}t=v_{0, car}t+\dfrac{1}{2}a_{car}t-7.5,$$ $$30t=45t+\dfrac{1}{2}\cdot15t^2-7.5,$$ $$7.5t^2+15t-7.5=0.$$

This quadratic equation has two roots: $t_1=0.41$ and $t_2=-2.41.$ Since time can't be negative the correct answer is $t=0.41\ h.$

Substituting $t$ into the kinematic equation for the scooter we find the distance (of the scooter) when the car catches him:

$$d_{sc}=v_{0,sc}t=30\ \dfrac{km}{h}\cdot0.41\ h=12.3\ km.$$