NUMERICAL PROBLEM BASED ON MOTION WITH UNIFORM ACCELERATION
Asked by | 1st Dec, 2008, 05:30: PM
The first quarter S = 100m.
S=ut+1/2 at2 : 100 = 1/2 x a x (10)2
We find that the acceleration produced during the first 100 m = 2m/s2 .
The final velocity at the end of the 10th sec is : v= u +at →v=2x10 = 20m/s.
The rest 300 m is travelled at the speed of 20m/s.Hence the time taken is 300/20 = 15s.
Hence the total time taken is 10+15 = 25sec.
Answered by | 2nd Dec, 2008, 08:34: PM
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