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?
please, explain your answer..
-thanks in advance.

Answer 1

You need to set their position functions equal to one another and so for the time t when that is true. That is when the tiger and the deer are in the same place meaning the tiger catches the dear
Xdear= 2t+15 deer position function.
(I integrated the velocity function )
To get the Tigers position function you must integrate the acceleration twice. This becomes
Xtiger=t^2
Now t^2=2t+15
Time t is when the tiger catches the deer
t^2-2t-15=0
(t-5)(t+3)=0 factored
t=5s is the answer you use (t=-3 is a meaningless solution)