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?
The data given in this numerical problem are in different units. So, we should first convert km h-1 unit into m s-1 unit.
For first car:
Initial velocity u = 52 km h-1
= \(\frac{52km}{1 h}\) = \(\frac{52\times 1000m}{1 \times 3600 s}\)
= 14.4 m s-1
Final velocity, v = 0 km h-1 = 0.0 m s-1
Time taken, t = 5 s
For second car:
Initial velocity, u = 3km h-1
= \(\frac{3km}{1 h}\) = \(\frac{3\times 1000 m}{1 \times 3600 s}\) = 0.8 m s-1
Final velocity, v = 0 km h-1
= 0.0 m s-1
Time taken, t = 10 s
The distance traveled by a moving body is given by the area under its speed-time graph. so, Distance traveled by the first car=Area of the triangle AOB
= 1/2 x OB x AO = 1/2 x 14.4 m s-1 x 5 s
= 1/2 x 14.4 x 5m = 36 m
Similarly, Distance travelled by the second car = Area of triangle COD
= 1/2 x OD x CO = 1/2 x 0.83m s-1 x 10 s
= 1/2 x 0.83 x 10 m = 4.1 m
Thus, the second car travels 4.1 m and the first car travels 36 m before coming to rest. So, the second car traveled farther after the brakes were applied.