A particle has an initial velocity of 9 m/s due east and a constant acceleration of 2 m/s2 due west. What is the distance covered by the particle in the fifth second of its motion ?
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9
Short answer:
s=ut+1/2at^2
s=9(5)+1/2(-2)(5)^2
s=45-25
s=20
Further explanation:
The particle is moving to east (right) with velocity 9m/s and a constant acceleration of 2m/s^2 due west which means deceleration with 2m/s^2.
Therefore a=-2.
s=ut+1/2at^2
s=9(5)+1/2(-2)(5)^2
s=45-25
s=20
Further explanation:
The particle is moving to east (right) with velocity 9m/s and a constant acceleration of 2m/s^2 due west which means deceleration with 2m/s^2.
Therefore a=-2.
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3
since we have acc on opposite direction so the particle will eventually get at rest... and that time is 4.5s... after that the particle will move is opposite direction in which it was going before...
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