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

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)#