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Ezample 32The distance travelled by a particle in time t is givenby(2-5 ms 3t Find (a the average speed of theparticle during the time 0 to 50s, and b) theinstantaneous speed at t-508 |
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Answer» ok given s = 2.5t² so, distance travelled by particle after 5secs is = 2.5*(5)² = 62.5m/s² now, average speed = total distance/time = 62.5/5 = 12.5 m/s and instantaneous speed at 5 sec is given by V = ds/dt = 5t = 5*5 = 25m/s..... but distance m/s² why? |
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