Question

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 1

Note: Refer (attachment).

As given in the figure above:

PR and SQ is the speed-time graph for given two cars with initial speeds:

$\implies$ 52 km/hr

$\implies$ 3 km/hr

Now:

First car A:

Initial Velocity (u):

$\implies$ 52 km/hr

$\implies$ $\frac{52 \times 5}{18}$

$\implies$ 14.4 m/s

Final velocity (v) = 0 m/s

Time (t) = 5 sec

Distance travelled by the first car A:

Before coming to rest = Area of Δ OPR

$\implies$ $\sf{ \frac{1}{2} \times OR \times OP}$

$\implies$ $\sf{ \frac{1}{2} \times 5 \times 52}$

$\implies$ $\sf{ \frac{1}{2} \times 5 \times (\frac{52 \times 1000}{3600} )}$

$\implies$ $\sf{ \frac{1}{2} \times 5 \times (\frac{130}{9} )}$

$\implies$ $\sf{ \frac{325}{9}}$

$\implies$ $\boxed{\sf{36.11 \: m}}$

Now:

For Car B:

Initial Velocity (u):

$\implies$ 3 km/hr

$\implies$ $\frac{3 \times 5}{18}$

$\implies$ 0.83 m/s

Final velocity (v) = 0 m/s

Time (t) = 10 sec

Now:

Distance travelled by the second car B:

Before coming to rest = Area of Δ OSQ

$\implies$ $\sf{\frac{1}{2} \times OQ \times OS}$

$\implies$ $\sf{\frac{1}{2} \times 10 \times 3}$

$\implies$ $\sf{ \frac{1}{2} \times 10 \times ( \frac{3 \times 1000}{3600} )}$

$\implies$ $\sf{ \frac{1}{2} \times 10 \times (\frac{5}{6} )}$

$\implies$ $\sf{ 5 \times (\frac{5}{6})}$

$\implies$ $\sf{\frac{25}{6}}$

$\implies$ $\boxed{\sf{4.16\:m}}$

Therefore:

Car B has travelled farther than Car A after the brakes were applied.

Answer 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 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.