Question

A particle is moving with uniform acceleration along a straight line ABC. Its velocity at 'A' and 'B' are 6m/s and 9m/s respectively. AB:BC=5:16 then its velocity at 'C' is

• A. 9.6m/s
• B. 12m/s
• C. 15m/s
• D. 21.5m/s

Answer 1

The correct option is C 15m/s

Given:

The velocity of particle at point $A$, $V_A=6m/s$

The velocity of particle at point $B$, $V_B=9m/s$

Acceleration of the particle is uniform

Ratio of distance between $AB$ and $BC$, $AB:BC=5:16$

To find:

The velocity of particle at $C$

Consider that the distance between $A$ and $B$ is $5x$ and between $B$ and $C$ is $16x$ as they in ratio of $5:16$

For motion between $A\to B$:

Using third equation motion : $v^2=u^2+2as$

final velocity is $V_B$ and initial is $V_A$ and displacement is $5x$

$(9{)}^2=(6{)}^2+2a\left(5x\right)$

$81=36+10ax$

$ax=\frac{45}{10}=4.5$

For motion between $B\to C$

$v^2=u^2+2a{s}_2$

$v^2=(9{)}^2+2a(16x)$

$v^2=\left(81\right)+32ax$

substitute $4.5$ for $ax$

$v^2=81+32\left(4.5\right)$

$=225$

$v=\sqrt{225}=15$ m/s