# Question #72a0e

Sep 12, 2016
1. Initial velocity$= {u}_{1}$
Acceleration ${a}_{1} = 2 m {s}^{-} 2$
Time of travel ${t}_{1} = 8 s$
Distance covered$= s$
Kinematic equations for car 1
$s = {u}_{1} {t}_{1} + \frac{1}{2} {a}_{1} {t}_{1}^{2}$
Inserting given values we have
$s = {u}_{1} \times 8 + \frac{1}{2} \times 2 \times {8}^{2}$
$\implies s = 8 {u}_{1} + 64$ .....(1)
2. Initial velocity$= {u}_{2}$
Acceleration ${a}_{2} = 9.5 m {s}^{-} 2$
Time of travel ${t}_{1} = 4 s$
Distance covered$= s$, as both cover the same distance.
Kinematic equations for car 2
$s = {u}_{2} {t}_{2} + \frac{1}{2} {a}_{2} {t}_{2}^{2}$
Inserting given values we have
$s = {u}_{2} \times 4 + \frac{1}{2} \times 9.5 \times {4}^{2}$
$\implies s = 4 {u}_{2} + 76$ .......(2)

Equating RHS of (1) with RHS of (2)
$8 {u}_{1} + 64 = 4 {u}_{2} + 76$, rearranging we get
$8 {u}_{1} = 4 {u}_{2} + 12$, dividing both sides with $4$
$2 {u}_{1} = {u}_{2} + 3$ .....(3)

Three situations arise:
(a) ${u}_{2} = {u}_{1}$
We get ${u}_{1} = {u}_{2} = 3 m {s}^{-} 1$. 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
${u}_{1} = \frac{{u}_{2} + 3}{2}$

(c) ${u}_{2}$ < ${u}_{1}$
from equation (3) we have
${u}_{2} = 2 {u}_{1} - 3$