a driver of a car travelling at 52 km//h applies the brakes and acceleration uniformly in the opposite direction. The car stops in 5 s. Another driver going at 34 km//h 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 ?
Answers
Explanation:
By the above data the graph is drawn and line m is car 1 and line n is car 2
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 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 traveling
at 52 kmph) traveled farther post the application of brakes.