A particle starting from a point $A$ and moving with a positive constant acceleration along a straight line reaches another point $B$ in time $T$ . Suppose that the initial velocity of the particle is $u > 0$ and $P$ is the mid-point of the line $A B$ . If the velocity of the particle at point $P$ is $v_{1}$ and if the velocity at time $\frac{T}{2}$ is $v_{2}$ , then

v_{1} = v_{2}

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v_{1} > v_{2}

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v_{1} < v_{2}

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v_{1} = \frac{1}{2} v_{2}

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Solution

The correct option is B $v_{1} > v_{2}$
R.E.F image
particle starts from A
we know $\to$
$V = u + a t$
$\Rightarrow V_{2} = u + a (\frac{T}{2}) . . . (i)$
Also, $V^{2} = u^{2} + 2 a s$
$\Rightarrow V_{1}^{2} = u^{2} + 2 a (\frac{A B}{2}) . . . (i i)$
Also, $S = u t + \frac{1}{2} a t^{2}$
$\Rightarrow (A B) = u T + \frac{1}{2} a (T^{2}) . . . (i i i)$
Using equation (ii) & (iii)
$\Rightarrow V_{1}^{2} = u^{2} + 2 a ⎛ ⎜ ⎜ ⎜ ⎝ \frac{u T + \frac{1}{2} a T^{2}}{2} ⎞ ⎟ ⎟ ⎟ ⎠$
$= u^{2} + a u T + \frac{1}{2} a^{2} T^{2}$
$= u^{2} + a u T + \frac{1}{4} a^{2} T^{2} + \frac{1}{4} a^{2} T^{2}$
$= (u + 2 \times u \times \frac{a T}{2} + \frac{a^{2} T^{2}}{4}) + \frac{1}{4} a^{2} T^{2}$
$= {(u + \frac{a t}{2})}^{2} \pm \frac{1}{4} a^{2} T^{2}$
$V_{1}^{2} = (v_{2})^{2} + \frac{1}{4} a^{2} T^{2}$
Clearly, $v_{1}^{2} > v_{2}^{2}$
as, $v_{1} > v_{2}$

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