A car with speed #v = "20 ms"^(-1)# passes a stopped police car. Right at this point, the car accelerates at #"2 ms"^-2# and the police car at #2a#. Show that, when the police car catches up to the other car, the distance is #d=(4v^2)/a#?
1 Answer
Mar 8, 2018
Let the police car catch up with the other car after time
Applicable kinematic equation is
#s=ut+1/2at^2#
-
For Police car
#d=0xxt+1/2(2a)t^2#
#=>d=at^2#
#=>t=sqrt(d/a)# .....(1) -
For car
#d=vt+1/2at^2#
Insert value of
#d=vsqrt(d/a)+1/2ad/a#
#=>d/2=vsqrt(d/a)#
Squaring both sides we get
#(d/2)^2=(vsqrt(d/a))^2#
#=>d^2=(4v^2)/ad#
#=>d^2-(4v^2)/ad=0#
#=>d(d-(4v^2)/a)=0#
Roots are
#d-(4v^2)/a=0#
#d=(4v^2)/a# .......(3)
Both roots are valid as these are values of
