A particle moving with a velocity equal to 0.4 m / s is subjected to an acceleration of 0.15 m / s 2 for 2 s in a direction at right angles to its direction of motion. The resultant velocity is

Question

A particle moving with a velocity equal to 0.4 m / s is subjected to an acceleration of 0.15 m / s 2 for 2 s in a direction at right angles to its direction of motion. The resultant velocity is (A) 0.7 m/s (B) 0.5 m/s (C) m/s (D) between 0.7 and 0.1 m/s

Tardigrade Admin · Accepted Answer

There will be no change in the velocity along horizontal direction, so

v_{x} = 0.4 m / s .

Velocity in perpendicular direction

v_{y} = 0 + 0.15 \times 2

= 0.3 m / s

So, resultant velocity,

v = v_{x}^{2} + y_{y}^{2}

= (0.4)^{2} + (0.3)^{2}

= 0.5 m / s

Q. A particle moving with a velocity equal to 0.4m/s is subjected to an acceleration of 0.15m/s2 for 2s in a direction at right angles to its direction of motion. The resultant velocity is

0.7 m/s

0.5 m/s

m/s

between 0.7 and 0.1 m/s

There will be no change in the velocity along horizontal direction, so vx​=0.4m/s. Velocity in perpendicular direction vy​=0+0.15×2 =0.3m/s So, resultant velocity, v=vx2​+yy2​​ =(0.4)2+(0.3)2​ =0.5m/s

Q. A particle moving with a velocity equal to $0.4 m / s$ is subjected to an acceleration of $0.15 m / s^{2}$ for $2 s$ in a direction at right angles to its direction of motion. The resultant velocity is

There will be no change in the velocity along horizontal direction, so
$v_{x} = 0.4 m / s .$
Velocity in perpendicular direction
$v_{y} = 0 + 0.15 \times 2$
$= 0.3 m / s$
So, resultant velocity,
$v = v_{x}^{2} + y_{y}^{2}$
$= (0.4)^{2} + (0.3)^{2}$
$= 0.5 m / s$