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A 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 s. Another driver going at 3 km h^–1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

Answer:

The speed v/s time graphs for the two cars can be plotted as follows.

The total displacement of each car can be obtained by calculating the area beneath the speed-time graph.

Therefore, displacement of the first car = area of triangle AOB

= (1/2)*(OB)*(OA)

But OB = 5 seconds and OA = 52 km.h^-1 = 14.44 m/s

Therefore, the area of the triangle AOB is given by: (1/2)*(5s)*(14.44ms^-1) = 36 meters

Now, the displacement of the second car is given by the area of the triangle COD

= (1/2)*(OD)*(OC)

But OC = 10 seconds and OC = 3km.h^-1 = 0.83 m/s

Therefore, area of triangle COD = (1/2)*(10s)*(0.83ms^-1) = 4.15 meters

Therefore, the first car is displaced by 36 meters whereas the second car is displaced by 4.15 meters.

**Therefore, the first car (which was travelling at 52 kmph) travelled farther post the application of brakes.**

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