Question

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 $AB$. 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

• A. $v_1=v_2$
• B. $v_1>v_2$
• C. $v_1<v_2$
• D. $v_1=\frac{1}{2}{v}_2$

Answer 1

The correct option is B $v_1>v_2$

R.E.F image

particle starts from A

we know $\to$

$V=u+at$

$\Rightarrow V_2=u+a\left(\frac{T}{2}\right)...\left(i\right)$

Also, $V^2=u^2+2as$

$\Rightarrow V_1^2=u^2+2a\left(\frac{AB}{2}\right)...\left(ii\right)$

Also, $S=ut+\frac{1}{2}a{t}^2$

$\Rightarrow \left(AB\right)=uT+\frac{1}{2}a\left(T^2\right)...\left(iii\right)$

Using equation (ii) & (iii)

$\Rightarrow V_1^2=u^2+2a\left(\frac{uT+\frac{1}{2}a{T}^2}{2}\right)$

$=u^2+auT+\frac{1}{2}{a}^2{T}^2$

$=u^2+auT+\frac{1}{4}{a}^2{T}^2+\frac{1}{4}{a}^2{T}^2$

$=(u+2\times u\times \frac{aT}{2}+\frac{a^2{T}^2}{4})+\frac{1}{4}{a}^2{T}^2$

$={(u+\frac{at}{2})}^2\pm \frac{1}{4}{a}^2{T}^2$

$V_1^2=(v_2{)}^2+\frac{1}{4}{a}^2{T}^2$

Clearly, $v_1^2>v_2^2$

as, $v_1>v_2$