Question #72a0e

1 Answer
Sep 12, 2016
  1. Initial velocity#=u_1#
    Acceleration #a_1=2ms^-2#
    Time of travel #t_1=8s#
    Distance covered#=s#
    Kinematic equations for car 1
    #s=u_1t_1+1/2a_1t_1^2#
    Inserting given values we have
    #s=u_1xx8+1/2xx 2xx8^2#
    #=>s=8u_1+64# .....(1)
  2. Initial velocity#=u_2#
    Acceleration #a_2=9.5ms^-2#
    Time of travel #t_1=4s#
    Distance covered#=s#, as both cover the same distance.
    Kinematic equations for car 2
    #s=u_2t_2+1/2a_2t_2^2#
    Inserting given values we have
    #s=u_2xx4+1/2xx 9.5xx4^2#
    #=>s=4u_2+76# .......(2)

Equating RHS of (1) with RHS of (2)
#8u_1+64=4u_2+76#, rearranging we get
#8u_1=4u_2+12#, dividing both sides with #4#
#2u_1=u_2+3# .....(3)

Three situations arise:
(a) #u_2=u_1#
We get #u_1=u_2=3ms^-1#. it is a valid solution.

From (a) itself we infer that solution exists even if #u_2=u_1#
As such the stated condition which is required to be proved is false.

(b) #u_2>u_1#
From (3) we have
#u_1=(u_2+3)/2#

(c) #u_2# < # u_1#
from equation (3) we have
#u_2=2u_1-3#