A particle starts from origin at t=0 with a velocity 5 hati m / s and moves in x-y plane under action of a force which produces a constant acceleration of (3 hati+2 hatj) m / s 2 The y -coordinate of the particle, when its x -coordinate is 84 m is

Question

A particle starts from origin at t=0 with a velocity 5 hati m / s and moves in x-y plane under action of a force which produces a constant acceleration of (3 hati+2 hatj) m / s 2 The y -coordinate of the particle, when its x -coordinate is 84 m is (A) 12 m (B) 24 m (C) 36 m (D) 48 m

Tardigrade Admin · Accepted Answer

The position of the particle is given by s=(ux+uy)+(1/2)(ax+ay) t2 Hence, x=5 t+1.5 t2, y=1 t2 Given x(t)=84 m , y= ? 5 t+1.5 t2=84 ⇒ t=6 s At t=6 s , y=62=36 m

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

$12 m$

$24 m$

$36 m$

$48 m$

The position of the particle is given by
$s = (u_{x} + u_{y}) + \frac{1}{2} (a_{x} + a_{y}) t^{2}$
Hence, $x = 5 t + 1.5 t^{2}, y = 1 t^{2}$
Given $x (t) = 84 m, y =$ ?
$5 t + 1.5 t^{2} = 84$
$\Rightarrow t = 6 s$
At $t = 6 s, y = 6^{2} = 36 m$

Q. A particle starts from origin at t=0 with a velocity 5i^m/s and moves in x−y plane under action of a force which produces a constant acceleration of (3i^+2j^​)m/s2 The y -coordinate of the particle, when its x -coordinate is 84m is

12m

24m

36m

48m