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A car moving towards north at a speed of 10 m/sturns right within 10 s maintaining the same speed.Average acceleration of the car during this timeinterval isAnswer is root 2 m/s2 |
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Answer» As the car turns, ACCELERATION is produced. So, initial velocity DUE NORTH (u)=10 m/s Final velocity east (V)=10 m/s Time interval (t)=10 s Following vectors:- √10²+10²=√200=10√2 m/s=change in velocity So, acceleration (a)=v-u/t =10√2/10 =√2 m/s So, AVERAGE acceleration is equal to √2 m/s |
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