Physics, asked by deeptibhatia15, 1 year ago

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

Answers

Answered by ggurmeetjeet03
4
Given : 
First Car  A: 
Initial Velocity= u= 52 km/hr=52x5/18=14.4 m/s
Final velocity= V= 0m/s
time =t=5sec
Refer attachment for Graph :
Distance travelled : Area of traingle AOB =1/2 OB x AO
=(1/2)x 14.4 x 5=72/2=36m
Second Car  B:
Initial Velocity= u= 3 km/hr=3x5/18=0.83m/s
Final velocity= V= 0m/s
time =t=10sec
Refer attachment for Graph :
Distance travelled : Area of traingle COD =1/2 OD x CO=(1/2)x 0.83 x 10=4.1m
So, after applying of brakes , Car A has travelled a distance of 3m and car B has travelled a distance of 4.1m.
∴ Car B has travelled farther than Car A after the brakes are applied.

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Answered by Anonymous
2
Hola user..............""

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QUESTION:- Driver of a car travelling at 52 km per hour applied the brakes and accelerates uniformly in the opposite direction the car stops in 5 seconds another driver going at 3 km per hour and another car applies is break slowly and stops in 10 seconds on the same graph paper brought the speed versus time graph for the two cars which of the two cars travel father after the brakes are applied


ANSWER:-


First of all we can write the given information

=> Car A......CASE1

=> U=52km/HR

=> 52x5/18

=> 14.4 m/s

Here V=0m/s

t= 5 sec

So, the distance traveled =area of AOB

1/2xOBxAO

=> 1/2x14.4x5

=> 72/2

=> 36 m

NOW


CarB..........CASE2

Given => U=3km/HR

=>3x5/18

=> 0.83m/s

Here also V=0m/s

T =10sec

Distance traveled =area of triangle COD

=> 1/2 X ODxCO

=> 1/2x0.83x10

=> 4.1 cm


car B had traveled farther than car A

AFTER APPLYING THE BREAKS

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