Question

Question #72a0e

Answer 1

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`