Question

A car, running at a speed of 72 km/h is a slowed down to 18 km/ h over a distance of 40 m. calculate the(1) retardation produced by its brakes, and(2) time for which brakes are applied​

Answer 1

Answer:

\begin{gathered}u = 72km {h}^{ - 1} \\ v = 18\: km {h}^{ - 1} \\ s = 40m = .04km \\ {v}^{2} = {u}^{2} - 2as \\ {18}^{2} = {72}^{2} - 2a \times .04 \\ 324 = 5184 - 0.08a \\ 0.08a = 5184 - 324 \\ a = \frac{4860}{0.08} \\ a = 60750 \: km {h}^{ - 2} \\ = \frac{60750 \times 1000}{60 \times 60 \times 60 \times 60} = 4.68 \: m {sec}^{ - 2} \\ \: \\ v = u - at \\ 18 = 72 - 60750t \\ t = \frac{72 - 18}{60750} = \frac{54}{60750} h\\ \frac{54 \times 60 \times 60}{60750} sec= 3.2 \: sec\end{gathered}

u=72kmh

−1

v=18kmh

−1

s=40m=.04km

v

2

=u

2

−2as

18

2

=72

2

−2a×.04

324=5184−0.08a

0.08a=5184−324

a=

0.08

4860

a=60750kmh

−2

=

60×60×60×60

60750×1000

=4.68msec

−2

v=u−at

18=72−60750t

t=

60750

72−18

=

60750

54

h

60750

54×60×60

sec=3.2sec

Explanation:

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