Physics, asked by robinkumar6230, 7 months ago

Driver of a car travelling at 52 km/h^-1 applies the brakes and accelerates uniformly in the opposite direction. the car stops in 5 seconds .another driver going at 3 kilometre per hour in another car applies is break slowly and stop in 10 seconds. on the same graph paper clock speed versus time graph for the two cars. which of the car travel father after the brakes are applied?

Answers

Answered by kumar6171
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Answered by sangeetadas59023
3

Answer:

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.

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

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

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

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

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

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

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

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 mDistance Travelled by second car before coming to rest =Area of Δ OSQ

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