Question

A train started from rest and accelerates uniformly at the rate of 2 metre per second for 10second . Then it's maintains a constant speed for 200 seconds. the breaks are then applied and the train is uniformly retarded and comes to rest in 15 seconds . draw a velocity versus graph and use it to find (1) the maximum velocity reached (2) the retardation in last 50 seconds (3)the total distance travelled and (4)the average velocity of the train.



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Answer 1

Answer:

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Answer 2

Answer:

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

Hope it was helpful.

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