Question

A vehicle moving with a constant acceleration from $A$ to $B$ in a straight line $AB$, has velocities $u$ and $v$ at $A$ and $B$ respectively. $C$ is the mid point of $AB$. If time taken to travel from $A$ to $C$ is twice the time to travel from $C$ to $B$ then the velocity of the vehicle $v$ at $B$ is:

• A. $5u$
• B. $6u$
• C. $7u$
• D. $8u$

Answer 1

The correct option is C $7u$

Let the velocity at A be $u$ and acceleration be $a$.

Thus . using the third equation of motion , $v^2=u^2+2as$

where $s$ is the distance between A and B.

From the first equation of motion,

Time taken to travel from A to C is $\frac{v^{\prime }-u}{a}$, where $v^{\prime }$ is the velocity of vehicle at $C$.

Similarly time taken to travel from C to B is $\frac{v-v^{\prime }}{a}$

Thus $\frac{v^{\prime }-u}{a}=2\frac{v-v^{\prime }}{a}$

$⟹v^{\prime }=\frac{u+2v}{3}$

Also $v^{\prime 2}-u^2=2a\frac{s}{2}=as$

Thus $v^2=u^2+2(v^{\prime 2}-u^2)=2v^{\prime 2}-u^2$

$=2(\frac{u+2v}{3}{)}^2-u^2$

On simplification, we get,

$v^2-8uv+7u^2=0$

$v^2-uv-7uv+7u^2=0$

$(v-u)(v-7u)=0$

$v=u,7u$

But , since the body is moving with constant acceleration , $v>u$

$⟹v=7u$