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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

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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