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.
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)
