# A particle moving at a velocity of 9.1m/s in the positive x direction is given an acceleration of 5m/s^2 in the positive y direction for 6s. What is the final speed of the particle? Answer in...

A particle moving at a velocity of 9.1m/s in the positive x direction is given an acceleration of 5m/s^2 in the positive y direction for 6s. What is the final speed of the particle? Answer in units of m/s

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### 2 Answers

The particle has an initial velocity of 9.1 m/s in the positive x-direction. It is accelerated at 5 m/s^2 in the positive y-direction for 6 s. Due to the acceleration the final velocity of the particle has a component in the direction of the positive x-axis of 9.1 m/s and a component in the direction of the positive y-axis of 5*6 = 30 m/s.

The final speed of the particle is the magnitude of its final velocity. This is the sum of the components of its velocity in the x and y axes which is equal to sqrt(30^2 + 9.1^2) = sqrt(900 + 82.81) = sqrt(982.81) = 31.34 m/s

The final speed of the particle is 31.34 m/s

Velocity of particle in x(+ve) direction = 9.1 m/s

Initial velosity of particle in y direction = 0 ......say u

Acceleration given in y direction = 5 m/s^2 ....say a

Time for which the acceleration works = 6 s ....say t

Velocity of particle in y direction after 6 seconds = u + a*t

= 6 *5 = 30 m/s

Final speed = magnitude of resultant velocity irrespective of the direction and resultant velocity = square root(9.1^2 + 30^2)

= square root(82.81+900) = 31.35 m/s

**The final speed **of the particle after 6 seconds **= 31.35 m/s**