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#?
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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
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