Question

A train starts from the rest and accelerates uniformly to a velocity 20 m/s in 10 minutes. It then travels

with constant velocity for 20 minutes. The breaks are then applied and the train is uniformly retarded.It

comes to rest in 5 minutes. Draw a velocity-time graph for the entire motion of train and find the total

distance travelled by the train using it​

Answer 1

Answer:17.69 m/s

Explanation:

Total distance covered(s) = Distance during acceleration(s1) +  distance during uniform motion(s2) + distance during retardation(s3)

For acceleration,

v=u+a∗t

vmax=2×10=20 m/s

s=ut+(1/2)at2

s1=(1/2)×2×102

    =100 m

For uniform motion,

s=vt

s2=20×200

    =4000 m

During retardation,

v=u+at

0=20+50t

  =−0.4 m/s2

s3=ut+(1/2)at2

    =20×50+(1/2)×(−0.4)×502

    =500 m

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

Total time taken, t=t1+t2+t3=260 s

Average velocity, vavg=Total TimeTotal Distance

                                     =4600/260=17.69 m/s