A car starts from rest and moves such that it | vedclass

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(B) The distance covered in a specific time interval is given by the product of the average speed during that interval and the duration of the interva

Vedclass · Accepted Answer

$30$

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(B) The distance covered in a specific time interval is given by the product of the average speed during that interval and the duration of the interval.Distance in the first second $(S_{1})$:$S_{1} = v_{av1} 	imes t_{1} = 5 \, ms^{-1} 	imes 1 \, s = 5 \, m$Distance in the second second $(S_{2})$:$S_{2} = v_{av2} 	imes t_{2} = 10 \, ms^{-1} 	imes 1 \, s = 10 \, m$Distance in the third second $(S_{3})$:$S_{3} = v_{av3} 	imes t_{3} = 15 \, ms^{-1} 	imes 1 \, s = 15 \, m$Total distance covered $(S_{T})$:$S_{T} = S_{1} + S_{2} + S_{3} = 5 + 10 + 15 = 30 \, m$

$A$ car starts from rest and moves such that it possesses average speeds of $5 \, ms^{-1}$,$10 \, ms^{-1}$,and $15 \, ms^{-1}$ in the first,second,and third seconds,respectively. What is the total distance covered by the car in these three seconds? ......... $m$

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