Collision Course

A car is moving with a velocity of 2 m/s 2\text{ m/s} . At a distance of 16 m 16 \text{ m} from it, a truck was moving with a constant velocity of 8 m/s 8 \text{ m/s} . The car driver immediately accelerated his car for a constant acceleration of 2 m/s 2 2\text{ m/s}^2 . After what time (in seconds) will the car strike the truck, if the truck and the car, none of them change their directions before collision?

This question had come to my mind when I was in the ninth standard though.


The answer is 8.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

It is pretty evident that before the car collides with the truck, the truck will cover some distance x x . Then clearly, the car before collision would cover a distance of ( 16 + x ) m (16+x)m by that time. Since the truck has a constant velocity, its distance x x is equal to 8 t 8t .

Secondly, the time taken by the car to collide with the truck is exactly equal to the time taken by the truck to cover x x , i.e., their reaction time is equal. Therefore, let the time t t be equal for both their distances. Now, the distance covered by the car exactly before collision:

After solving the quadratic equation, we find that the only legitimate and permissible value of t t is 8 8 (the another value is negative and time cannot be negative). Therefore, required time is 8 8 seconds.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...