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 #t# after it starts.
Applicable kinematic equation is

#s=ut+1/2at^2#

  1. For Police car
    #d=0xxt+1/2(2a)t^2#
    #=>d=at^2#
    #=>t=sqrt(d/a)# .....(1)

  2. For car

    #d=vt+1/2at^2#

Insert value of #t# from (1)

#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=0and #

#d-(4v^2)/a=0#
#d=(4v^2)/a# .......(3)

Both roots are valid as these are values of #d# when cars meet. However, catching up after acceleration is given by (3).

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