Question

a driver of a car travelling at 52 km/hr applies brakes and accelerates uniformly in the opposite direction.The car stop at in 5 s. another driving going at 3 km/h in another car applice his brakes slowly and Stop un 10s.On the same graph paper,plot the speed versus time graphs for the two cars.Which of the brakes were applied?​

Answer 1

Explanation :-

Given for first car:

Initial velocity,u => 52km/hr

$\sf{}\implies \dfrac{52\times 1000}{1\times 60\times 60}m/s$

$\sf{}\implies \dfrac{52000}{3600}m/s$

$\sf{}\therefore 14.44m/s$

Final velocity,v => 0m/s

Time taken ,t => 5s

Given for second car:

Initial velocity of the second car,u => 3km/hr

$\sf{}\implies \dfrac{3\times 1000}{1\times 60\times 60}m/s$

$\sf{}\implies \dfrac{3000}{3600}m/s$

$\sf{}\therefore 0.83m/s$

Final velocity of the second car,v => 0m/s

Time taken by second car to stop,t => 10s

Solution:

In the graph,blue slope shows the velocity of the first car and green slope shows the velocity of the second car.

Distance is calculated by the area under the slope of the graph.

Therefore,distance covered by 1st car = Area of ∆OAD

$\sf{}\implies Distance,s=\dfrac{1}{2}\times OD\times OA$

$\sf{}\implies \dfrac{1}{2}\times 14.4m/s\times 5s$

$\sf{}\implies 7.2m/s\times 5s$

$\sf{}\therefore 36m$

Thus,distance covered by second car = Area of ∆OBC

$\sf{}\implies Distance,s=\dfrac{1}{2}\times OC\times OB$

$\sf{}\implies \dfrac{1}{2}\times 9.4m/s\times 10s$

$\sf{}\implies 4.7/s\times 10s$

$\sf{}\therefore 47m$

Therefore,2nd car travelled father.

[Graph is in attachment]

Answer 2

Click the image up

Hope it will help you!!!!