Question

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

Answer 1

Answer:

The  resultant velocity is​ 0.5m/s

Explanation:

According to the problem the particle is moving with the velocity  0.4 m/s

Now the acceleration of the particle is 0.15 m/s^2 for the time 2 s

Therefore the velocity along with y component is

V(y) = u(y)+at  

V(y) = 0 + (0.15)2  

V(y) = 0.3 m/s

The velocity along with x component is

V(x)=0.4 m/s

Therefore the resultant velocity is

V(r) = √ (0.4^2 + 0.3^2)

    = 0.5m/s

Answer 2

Answer:

0.5 m/s  ( angle = 37°)

0.5 m/s  ( angle = 37°)

Explanation:

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

Velocity = 0.4 m/s

acceleration of 0.15 m/s² at 90° angle

Initially Velocity at right angle = 0 m/s

t = 2sec

Velocity at right angle = 0 + 0.15*2 =  0.3 m/s

acceleration in direction of motion = 0 m/s

Velocity in direction of motion = 0.4 m/s at t = 2

Velocity at t - 2sec = √(0.4)² +(0.3)²  =  √0.25 = 0.5 m/s

Angle = α => Tanα = 0.3/0.4 => Tanα = 0.75 => α = 36.87°

=> α = 37° (approx)

Resultant Velocity = 0.5 m/s at angle of 37°  to initial Direction of motion