Question

a train start from rest and accelerates uniformly at 100m - minute ^_2 for 10 minutes. It then , maintains a constant velocity for 20 minutes. The brakes are then applied and the train is uniformly retarded and comes to rest in the 5minutes. Draw a velocity - time graph and use it to find: (¡) the maximum velocity reached , (¡¡) the retardation in the last minutes, (¡¡¡) today distance travelled, and (¡v) the average velocity of the train​

Answer 1

Answer:

in which class are you

I am in 9th

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