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A driver of a car A travelling at $52 km / h$ applies the brakes and accelerates uniformly in the opposite direction. The car stops in $5 s$ . Another driver B going at $30 km / h$ in another car applies his brakes slowly and stops in $10 s$ . Which car travelled farther and at what distance?

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Car B,

41.5 m

Car B,

52 m

Car A,

46.2 m

Car A,

44.8 m

answer is A.

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Detailed Solution

Car B will travel farthest at

41.5 m

distance.
Given:
For the first car A,
Initial speed,

u_{1} = 52 km / h

\Rightarrow = \frac{52 \times 100}{60 \times 60}

\Rightarrow = 14.4 m / s

The final velocity,

v_{1} = 0

in time

t = 5 \sec

.
For second car B,
Initial speed,

u_{2} = 30 km / h

= \frac{30 \times 1000}{60 \times 60}

= 8.3 m / s

The final velocity,

v_{2} = 0

in time

t = 10 \sec

.
For car A
we know

v = u + at

\Rightarrow o = 14.4 + a \times 5

a = \frac{- 14.4}{5} m / s^{2}

We know

s = ut + \frac{1}{2} a t^{2}

\Rightarrow s = 14.4 \times 5 + \frac{1}{2} (\frac{- 14.4}{5}

)

\times 5^{2}

\Rightarrow s = 72 - 36

= 36 m

For car B
we know

v = u + at

\Rightarrow

o = 8.3 + a \times 10

\frac{- 8.3}{10} m / s^{2} = a

We know

s = ut + \frac{1}{2} a t^{2}

\Rightarrow s = 8.3 \times 10 + \frac{1}{2} (\frac{- 8.3}{10}

)

\times 10^{2}

\Rightarrow s = 83 - 41.5

\Rightarrow

s = 41.5 m

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