Question

A train starts from rest and accelerates uniformly at a rate of 2 m s ^-2 for 15 seconds .It then maintains a constant velocity for 300 seconds .It then experiences a uniform retardation when brakes are applied and comes to rest in 50 seconds . find the maximum velocity reached.​

Answer 1

Answer:

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

Answer 2

Answer:

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