Question

18. An express train starts from rest and accelerates
uniformly at a rate of 1 m s2 for 20 s. It then maintains
a constant speed for 120 s. Then, the driver applies
brakes and the train comes to rest in 10 s. Calculate
the (a) maximum velocity of the train, (b) retardation
on applying brakes, and (c) total distance travelled by
the train.


I need the answer asap please , with full process and formula .​

Answer 1

Answer:

Correct option is

C

17.69ms

−1

Total distance covered(s) = Distance during acceleration(s

1

) + distance during uniform motion(s

2

) + distance during retardation(s

3

)

For acceleration,

v=u+a∗t

v

max

=2×10=20 m/s

s=ut+(1/2)at

2

s

1

=(1/2)×2×10

2

=100 m

For uniform motion,

s=vt

s

2

=20×200

=4000 m

During retardation,

v=u+at

0=20+50t

=−0.4 m/s

2

s

3

=ut+(1/2)at

2

=20×50+(1/2)×(−0.4)×50

2

=500 m

Total distance travelled, s=s1+s2+s3=100+4000+500=4600 m

Total time taken, t=t

1

+t

2

+t

3

=260 s

Average velocity, v

avg

=

Total Time

Total Distance

=4600/260=17.69 m/s

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