Physics, asked by chaitanayasharka, 1 year ago

a driver of car travelling at 2:00 p.m. people are applied the brakes and accelerates uniformly in the opposite direction the car stops in 5 second on another driver going at 34 kilometre per hour in another car applies is break slowly and stops car stops in 10 second on the same graph plot the time versus time speed versus time graph for the two gases which of the two cases driver father after the break replied

Answers

Answered by sajanpreet2
1
very big question very big answer.....
Answered by sangeetadas59023
4

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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