Question

A Train on a straight railway track starts from rest and accelerates at 2.0 m/s2 in 10sec. Then the train covers a distance of 300 m at constant speed for 20 sec until the brakes are applied, stopping the train in a uniform manner in an additional 5.0 s. (a) How far is the train in first 30sec? (b) What is the average velocity of the train for the complete journey?

Answer 1

(A)

Total distance covered (d) in first 30 sec= Distance during acceleration(s1​) + distance during uniform motion(s2​)

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​=20×20

=400 m

During retardation,

v=u+at

0=20+a(5s)

a=−4 m/s²

s3​=ut+(1/2)at²

=20×5+(1/2)×(−4)×5²

=50 m

Total distance travelled,

d=s1+s2=100+400=500 m

(B)

Total distance , d=s1+s2+s3=500+50=550 m

Total time taken,

t=t1​+t2​+t3​=35 s

Average velocity,

v=Total Distance​/(Total time)

=550/35

=15.7 m/s