Question

A train starts its journey from station p.
accelerates at the rate of 2 m/s2, and
reaches its maximum speed in 10 s. It
maintains this speed for 30 min and retards
uniformly to rest at the station Q after the
next 20 s. Calculate:
(a) The maximum speed of the train.
(b) Retardation
(c) The distance between stations P and Q.
​

Answer 1

Answer:

(a) u = 0 m/s

a = 2m/s^2

t = 10 s

v

= u + at

= 0 + 2×10 m/s

= 20 m/s(Answer)

(b) u = 20m/s

t = 20s

v = 0m/s

a

= (v-u)/t

= (0 - 20)÷20 m/s^2

= -1 m/s^2

r = -a = 1 m/s^2(Answer)

(c) s1 = ut + 0.5×a×t^2 = 0+0.5×2×10×10 m = 100 m

s2 = vt = 20×30×60 m = 36000 m

s3 = ut - 0.5×r×t^2 = 20×20 - 0.5×1×20×20 m = 200 m

Total distance

= s1 + s2 + s3

= 100 + 200 + 36000 m

= 36.3 Km (Answer)