First slide

Rectilinear Motion

Question

A particle moves with uniform acceleration along a straight line AB. Its velocities at A and B are 2 m/s and 14 m/s, respectively. M is the mid-point of AB. The particle takes t₁ seconds to go from A to M and t₂ seconds to go from M to B. Then t₂/t₁ is

Moderate

A

1:1
B

2:1
C

1:2
D

3:1

Solution

$v^{2} = 2^{2} + \frac{2 a s}{2}$
$\begin{array}{l} \Rightarrow v^{2} - 4 = a s . . . . . (i) \\ 14^{2} - v^{2} = \frac{2 a s}{2} \\ \Rightarrow 196 - v^{2} = a s . . . . . (i i) \end{array}$
From (i) and (ii),
$\begin{array}{l} v^{2} - 4 = 196 - v^{2} \Rightarrow v = 10 m / s \\ Now t_{1} = \frac{v - 2}{a} = \frac{10 - 2}{a} = \frac{8}{a} \\ t_{2} = \frac{14 - v}{a} = \frac{14 - 10}{a} = \frac{4}{a} \Rightarrow \frac{t_{2}}{t_{1}} = \frac{4 / a}{8 / a} = \frac{1}{2} \end{array}$

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