Question

A tiger started running when it saw a deer running at uniform velocity 2m/s at 15m far from it ..if the tiger ran at acceleration 2m/s^2..when and where did the tiger catch the deer?

Answer 1

The position x for the tiger with a constant acceleration a is given by:

`x = (1)/(2) at^2 + v_0t + x_0, v_0 = 0, x_0 = 0`

The position x for the deer with constant velocity is given by:

`x = vt + x_0, x_0 = 15m`

When the position of the tiger and the deer are the same the time t will be:

`(1)/(2) at^2 = vt + v_0, a = 2 (m)/(s^2),v = 2 (m)/(s) \\ t^2 = 2t + 15 \\ (t - 2)t = 15 \\ t = 5s`
At t = 5s the position of the tiger and the deer are:

`x = (1)/(2) at^2 = (1)/(2)2 (m)/(s) (5s)^2 = 25m \\ \\ x = vt + v_0 = 10m + 15m = 25m`