# A vehicle moving with constant acceleration from A to B in a straight line AB, has velocities \[u\] and $v$ at A and B respectively. C is the midpoint of AB. If the 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. $8u$

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**Hint:**When acceleration is constant in a straight line motion, time taken to reach from initial to final point is given by $t = \dfrac{{v - u}}{a}$ where $v$ is the final velocity, \[u\] is the initial velocity and $a$ is the acceleration.

We can also apply the formula ${v^2} - {u^2} = 2as$ where $v$ is the final velocity, \[u\] is the initial velocity, $a$ is the acceleration and $s$ is the displacement from initial to final point.

**Complete step by step solution:**

As given in the question the vehicle has velocities \[u\] and $v$ at $A$ and $B$ respectively and the time taken to travel from $A$ to $C$ is twice the time to travel from $C$ to $B$. The acceleration is constant throughout the motion.

So, we will first find the relation between the time taken to travel from $A$ to $C$ and time to travel from $C$ to $B$.

Let the velocity of the vehicle be ${v_c}$ at point $C$, the constant acceleration throughout the motion be $a$ and total displacement AB is $s$.

As we know, when acceleration is constant in a straight line motion, time taken to reach from initial to the final point is given by $t = \dfrac{{v - u}}{a}$ where $v$ is the final velocity, \[u\] is the initial velocity, and $a$ is the acceleration.

So, time taken by the vehicle from $A$ to $C$, ${t_{AC}} = \dfrac{{{v_c} - u}}{a}$ and time taken by the vehicle from $C$ to $B$, ${t_{CB}} = \dfrac{{v - {v_c}}}{a}$ .

Now as given in the question, ${t_{AC}} = 2{t_{CB}}$

Which is written as, $\dfrac{{{v_c} - u}}{a} = 2\dfrac{{\left( {v - {v_c}} \right)}}{a}$

On solving the equation further, ${v_c} - u = 2v - 2{v_c}$

On simplifying we have, ${v_c} = \dfrac{{2v + u}}{3}$ …(i)

Now, we can also apply the formula ${v^2} - {u^2} = 2as$ where $v$ is the final velocity, \[u\] is the initial velocity , $a$ is the acceleration and $s$ is the displacement from initial to final point.

As, $C$ is the mid point of $AB$, so, $AC = \dfrac{s}{2}$ .

So, from $A$ to $C$, $v_c^2 - {u^2} = 2 \times a \times \dfrac{s}{2} = as$ …(ii)

Now, from $A$ to $B$, ${v^2} - {u^2} = 2as$

Substituting the value of $as$ from equation (ii), we get,

${v^2} - {u^2} = 2\left( {v_c^2 - {u^2}} \right) = 2v_c^2 - 2{u^2}$

On solving further we have, ${v^2} = 2v_c^2 - {u^2}$

Now, substituting the ${v_c}$ from equation (i) we have,

\[{v^2} = 2{\left( {\dfrac{{2v + u}}{3}} \right)^2} - {u^2}\]

On simplification we get,

$

{v^2} - 8uv + 7{u^2} = 0 \\

{v^2} - uv - 7uv + 7{u^2} = 0 \\

$

On further solving the equation we have,

$\left( {v - u} \right)\left( {v - 7u} \right) = 0$

So, $v = u,7u$

But $v > u$ (as vehicle is in constant acceleration)

So, $v = 7u$

**$\therefore$The required velocity of the vehicle is $7u$. Hence, option (C) is the correct answer.**

**Note:**

Carefully substitute the initial and final velocities in the equations $t = \dfrac{{v - u}}{a}$ and ${v^2} - {u^2} = 2as$ and always remember that these formulae are applicable only when the particle is moving with a constant acceleration in a straight line.