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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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...
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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
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