Question

ii). A car moving along a straight road with a
speed of 120 km/hr, is brought to rest by
applying brakes. The car covers a distance
of 100 m before it stops. Calculate (i) the
average retardation of the car (ii) time
taken by the car to come to rest.
[Ans: 50/9 m/sec, 6 sec)​

Answer 1

Answer:

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Explanation:

given that,

Initial speed of the car ,u = 120 km/hr = 33.33 m/s

Finally it comes to rest, v = 0

Distance ,s = 100m

a) Let a is the average retardation, Using third equation of motion as:

$v {}^{2} - u {}^{2} = 2as \\ \\a = \frac{ { - u}^{2} }{2s} \\ \\ a = \frac{ { - (33.33)}^{2} }{2 \times 100} \\ \\ a = - 5.55m/s {}^{2}$

b) Leti is the time taken by the car to come to rest. It is given by:

$t = \frac{v - u}{a} \\ \\ t = \frac{ - 33.33}{ - 5.55} \\ \\ t = 6 \: sec$

Hence, this is the required equation

Answer 2

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Explanation:

given that,

Initial speed of the car ,u = 120 km/hr = 33.33 m/s

Finally it comes to rest, v = 0

Distance ,s = 100m

a) Let a is the average retardation, Using third equation of motion as:

$\begin{gathered}v {}^{2} - u {}^{2} = 2as \\ \\a = \frac{ { - u}^{2} }{2s} \\ \\ a = \frac{ { - (33.33)}^{2} }{2 \times 100} \\ \\ a = - 5.55m/s {}^{2} \end{gathered}$

b) Leti is the time taken by the car to come to rest. It is given by:

$\begin{gathered}t = \frac{v - u}{a} \\ \\ t = \frac{ - 33.33}{ - 5.55} \\ \\ t = 6 \: sec\end{gathered}$

Hence, this is the required equation

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