Question

A particle starts from origin at t = 0 with a velocity $5\widehat{i}ms^{- 1}$and moves in x-y plane under the action of force which produces a constant acceleration of $3\widehat{i} + 2\widehat{j}ms^{- 2}$. The y-coordinate of the particle at the instant when its x-coordinate is 84 m is

• A. 12 m
• B. 24 m
• C. 36 m
• D. 48 m

Answer 1

Answer: (C)

36 m

Answer 2

The positions of the particle is given by

$\overset{\rightarrow}{r} = \overset{\rightarrow}{r_{0}} + {\overset{\rightarrow}{v}}_{0}t + \frac{1}{2}\overset{\rightarrow}{a}t^{2}$

Where, ${\overset{\rightarrow}{r}}_{0}$is the position vector at t = 0

${\overset{\rightarrow}{v}}_{0}$is the velocity at t = 0

Here, ${\overset{\rightarrow}{r}}_{0} = 0,{\overset{\rightarrow}{v}}_{0} = 5\widehat{i}ms^{- 1},\overset{\rightarrow}{a} = 3\widehat{i} + 2\widehat{j}ms^{- 2}$

$\therefore\overset{\rightarrow}{r} = 5t\widehat{i} + \frac{1}{2}(3\widehat{i} + 2\widehat{j})t^{2} = (5t + 1.5t^{2})\widehat{i} + 1t^{2}\widehat{j}$

Compare it with $\overset{\rightarrow}{r} = x\widehat{i} + y\widehat{j},$ we get

$x = 5t + 1.5t^{2},y = 1t^{2}$

Given : x = 84 m

$\therefore 84 = 5t - 1.5t^{2}$

$1.5t^{2} + 5t - 84 = 0$

$3t^{2} + 10t - 168 = 0$

On solving, we get t = 6s

At t = 6s, y = (1) $(6)^{2} = 36m$

$\overset{\rightarrow}{r} = \overset{\rightarrow}{r_{0}} + {\overset{\rightarrow}{v}}_{0}t + \frac{1}{2}\overset{\rightarrow}{a}t^{2}$