Question

A train starts from rest with a constant acceleration 3 m/s2

for 5 seconds. Now it moves with

a constant speed for the next 300 seconds and finally retards to come to rest again in the next 15

seconds​

Answer 1

solution-;

• 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