A train starts from rest and is accelerated with an acceleration . At time it decelerates with deceleration and finally stops after travelling total distance of 300 m.Find the distance travelled by train in time
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Step-by-step 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
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