Question

A particle with initial velocity v = (-21+4)) m/s undergoes a constant acceleration a = 3 m s-² at 37° with positive direction of x-axis. What is particle's velocity at time t = 5 second?​

Answer 1

Answer:

Hey Mate!!

Explanation:

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Answer 2

Answer:

velocity of the particle( at t = 5 sec) = 14i + 13j m/s.

Explanation:

Given, velocity of the particle  V = 2i + 3j m/s.

Let's divide the motion of the particle into two parts, one along the X-axis other along Y-axis.

In X-axis :

velocity of the particle = 2 m/s.

and the acceleration = 3 cos 37° = 3 x 4/5 = 12/5 m/s².

And we know that v = u + at ( if acceleration is constant).

so the velocity along the X direction at (t = 5 sec) = 2 + 12 = 14 m/s.

In Y-axis :

velocity of the particle = 4 m/s.

and the acceleration = 3 sin 37° = 3 x 3/5 = 9/5 m/s².

so the velocity along the Y direction at (t = 5 sec) = 4 + 9 = 13 m/s.

Therefore, the final velocity of the particle = 14i + 13j m/s.