# A cheetah is pursuing an impala. The impala is running in a straight line constant speed of 16 m/s. The cheetah is 10 m behind the impala running at..The cheetah is 10 m behind the impala running...

A cheetah is pursuing an impala. The impala is running in a straight line constant speed of 16 m/s. The cheetah is 10 m behind the impala running at..

The cheetah is 10 m behind the impala running at 20 m/s but tiring, so that it is decelerating at 1 m/s^2. Find an expression of the gap between the cheetah and the impala t seconds later and show that it is 0.5t^2 -4t+10 m. Will the impala get away?

Use 1D kinemtaics equations to solve this question.

### 1 Answer | Add Yours

The initial speeds of the impala and the cheetah are 16m/s and 20 m/s.

The cheetah has a decileration of 1m/s^2. So the distance travelled by cheetah in t seconds is given by s(t) = ut+(1/2)at^2, where u is the initial speed , a is the acceleration (or decileration ) of the moving object. Applying this to the cheetah, u = 20m/s, a = -1m/s^2. So s(t) = 20t+(1/2)(-1)t^2. Or s(t) = 20t^2 -(1/2)t^2.

After t seconds the distance covered bt impala at a constant speed of 16m/s is 16t.

The initiial gap between impala and cheetah (cheetah is behind) is 10m

Therefore after t seconds the behind gap between the cheetah and the impala = 10+16t - [20t-(1/2)t^2] = 10 -4t+(1/2)t^2. So the cheetah is behind the impala by 0.5t^2 -4t+10 after t seconds.