A car slows down from 108 km/hr to 18 km/hr over a distance of 40 meter. If the brakes are applied with the same force . Calculate

1) total time in which car comes to rest .

2) total distance travelled by it.

### Asked by vishwadiprathod0070.9sdatl | 7th May, 2020, 11:10: AM

Expert Answer:

### When, u = 108 km/h = 30 m/s
and v = 18 km/h = 5 m/s
s = 40 m
We can calculate the acceleration using 3rd equation of motion: -
v^{2} - u^{2} = 2as
(5)^{2} - (30)^{2} = 2 a (40)
25 - 900 = 80a
a = - 10.93 m/s^{2}
Negative sign indicates that it is retardation.
**i)Total time in which car comes to rest: - **
** As the brakes are applied with same force to bring car to rest, acceleration remains same i.e. 10.93 m/s**^{2}.
v = 0, u = 30 m/s
a = -10.93 m/s^{2}
Thus,
v = u + at
0 = 30 + (-10.93) t
-30 = -10.93 t
Thus,
t = 2.74 s
**ii) Distance travelled by car during this time is given by, **
v^{2} - u^{2} = 2as
(0)^{2} - (30)^{2} = 2(-10.93) s
- 900 = 21.86 s
s = 41.17 m

^{2}- u

^{2}= 2as

^{2}- (30)

^{2}= 2 a (40)

^{2}

**i)Total time in which car comes to rest: -**

**As the brakes are applied with same force to bring car to rest, acceleration remains same i.e. 10.93 m/s**

^{2}.^{2}

**ii) Distance travelled by car during this time is given by,**

v

^{2}- u^{2}= 2as(0)

^{2}- (30)^{2}= 2(-10.93) s- 900 = 21.86 s

s = 41.17 m

### Answered by Shiwani Sawant | 7th May, 2020, 05:52: PM

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