A jeep chases a thief with a constant velocity v. When the jeep is at distance d from the thief, he starts to run with a constant acceleration a. Show that the police will be able to catch the thief when v >= sqrt(2ad)?

1 Answer

Shown

Explanation:

Given is speed of police jeep = v

Distance between the thief and the police jeep=d

At this point constant acceleration of thief = a.

Let the time when both meet =t
Let s be the distance covered by thief.
Using kinematic expression
s = ut+1/2 at^2
we get
s = 1/2 at^2 ......(1)
During this time interval distance traveled by police jeep = s + d
Which is also = vt

Equating two we get
s+d=vt
=>s = vt - d ......(2)
From (1) and (2) we get
1/2 at^2=vt - d
=>at^2 - 2vt + 2d = 0

Solving quadratic for t
t=(-b+-sqrt(b^2-4ac))/(2a)
t=(2v+-sqrt(4v^2-4axx2d))/(2a)
=>t=(v+-sqrt(v^2-2ad))/(a)
For a real time t, the discriminant must be >=0 and t can not be negative. From the first condition we get
v^2-2ad>=0
=>v^2>=2ad
=>v>=sqrt(2ad)