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.
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.