A driver of a car travelling at 54 km/h applies brakes and accelerates uniformly in the opposite direction. Car stops in 5 sec. Calculate the distance traveled after the breaks were applied.
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Answer:
hii mate
Step-by-step explanation:
As given in the figure below PR and SQ is the Speed-time graph for given two cars with initial speeds 52 km/hr and 3 km/hr respectively.
Distance Travelled by first car before coming to rest =Area of Δ OPR
= (1/2) x OR x OP
= (1/2) x 5 s x 52 kmh-1
= (1/2) x 5 x (52 x 1000) / 3600) m
= (1/2) x 5x (130 / 9) m
= 325 / 9 m
= 36.11 m
Distance Travelled by second car before coming to rest =Area of Δ OSQ
= (1/2) x OQ x OS
= (1/2) x 10 s x 3 kmh-1
= (1/2) x 10 x (3 x 1000) / 3600) m
= (1/2) x 10 x (5/6) m
= 5 x (5/6) m
= 25/6 m
= 4.16 m
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Answered by
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Answer:
ut+1/2×at^2
54km/hr=54×5/18m/s=15m/s
0×5+1/2×15×25s
=1/2×375m/s^2
375/2m
=187.5m
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