Question

A particle starts from origin at t = 0 with velocity of 15$\hat{i}$ and moves in x-y plane under the action of force which produce a constant acceleration of $15\hat{i}+20\hat{j}$. Find y-coordinate of particle when x-coordinate is 180 m.

Answer 1

x-coordinate :

Velocity $u_x=15m/s$

Acceleration $a_x=15m/s^2$

Using $S_x=u_{x}t+\frac{1}{2}{a}_x{t}^2$

Or $180=15t+\frac{1}{2}\times 15t^2$

$⟹t=4s$

y-coordinate :

Initial velocity $u_y=0$

Acceleration $a_y=20m/s^2$

$S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

Or $S_y=0+\frac{1}{2}\times 20\times 4^2=160m$