Question

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 1

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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Answer 2

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