Question

A particle moves in the X-Y plane with a constant acceleration of 1.5 $m{s}^{-2}$ in the direction making an angle of 37${}^0$ with the X-axis. At t=0 the particle is at the origin and its velocity is 8.0 m/s along the X-axis. Find the velocity of the particle at time t=4.0 sec

• A. 10.3 m/s
• B. 13.3 m/s
• C. 15.3 m/s
• D. 19.3 m/s

Answer 1

Answer: (B)

13.3 m/s

Acceleration of particle along x-axis=$a_x=1.5\cos {37}^{\circ }m/s^2=1.2m/s^2$

Acceleration of particle along y-axis=$a_y=1.5\sin {37}^{\circ }m/s^2=0.9m/s^2$

After 4 seconds, velocity of particle in x-axis=$v_0+a_{x}t$
$=12.8m/s$

velocity of particle in y-axis=$a_{y}t$
$=3.6m/s$

Hence net velocity=$\sqrt{v_x^2+v_y^2}$
$=13.3m/s$