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CBSE Class 9 Answered

In a test safety restraint a test car slowed from 50m/s to rest in a time of 0.4seconds (a) what was the acceleration (b) How far did the car travel

 

Asked by 8318813552s | 07 Feb, 2019, 11:09: PM
Expert Answer
To get acceleration, we need to use the relation " v = u+a×t ", where v is the final speed which is zero in this case,
a is acceleration and t is time to bring the vechile from initial speed u to rest.
 
hence,  0 = 50 + a×0.4   or  a = - 50/0.4 = - 125 m/s2
 
Formula to be used to get distance S  travelled  " v2 = u2+ 2aS "
 
hence,  0  = 50×50 - 2×125×S   or   S  = (50×50)/(2×125) = 10 m
Answered by Thiyagarajan K | 08 Feb, 2019, 10:00: AM
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