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?

Question

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Answer 1

(A) : Initial speed of the car A $u = 52$ $k m h^{- 1} = 52 \times \frac{5}{1 8} = 14.44$ $m s^{- 1}$
The car stops in 5 seconds i.e $v = 0$ at $t = 5$

(A) : Initial speed of the car A $u = 3$ $k m h^{- 1} = 3 \times \frac{5}{1 8} = 0.83$ $m s^{- 1}$
The car stops in 10 seconds i.e $v = 0$ at $t = 10$
With the help of these initial and final points for both the cases, we can plot the graph of speed vs time.
Area under the speed-time graph gives the distance covered.
$∴$ Distance covered by car A $x_{A} = \frac{1}{2} \times 14.44 \times 5 = 36.1$ m
$∴$ Distance covered by car A $x_{B} = \frac{1}{2} \times 0.83 \times 10 = 4.15$ m

Thus car A travels more distance than B.

Answer 2

(A) : Initial speed of the car A

u = 52

k m h^{- 1} = 52 \times \frac{5}{1 8} = 14.44

m s^{- 1}

The car stops in 5 seconds i.e

v = 0

at

t = 5

(A) : Initial speed of the car A

u = 3

k m h^{- 1} = 3 \times \frac{5}{1 8} = 0.83

m s^{- 1}

The car stops in 10 seconds i.e

v = 0

at

t = 10

With the help of these initial and final points for both the cases, we can plot the graph of speed vs time.

Area under the speed-time graph gives the distance covered.

∴

Distance covered by car A

x_{A} = \frac{1}{2} \times 14.44 \times 5 = 36.1

m

∴

Distance covered by car A

x_{B} = \frac{1}{2} \times 0.83 \times 10 = 4.15

m

Thus car A travels more distance than B.