Question

A train starts it's 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 after the next
20 s. Calculate:
(a) The maximum speed of the train.
(b) Retardation
(c) The distance between stations P and Q.
(Ans.(a) 72 km/h (b) 1 m/s? (c) 36.3 km)
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Answer 1

Answer:

Let's divide the entire journey into three parts,1.with uniform acceleration2.uniform speed3.uniform retardation

• a) Maximum speed of train is attained at 10th second (given)

therefore, we will have to see speed at 10th second,

V=at

V=2×10

v=20m/s

v=20×(18/5)km/hr

v=72km/hr

• b) retardation is the negative acceleration,

so, a=(v-u)t

here, v=final velocity=0 (since the body comes to rest)

u=initial velocity=20m/s (because,this is the maximum speed attained in the first region, which is maintained throughout the second region,so, at the beginning of the third region, body will have this velocity)

a=(0-20)/20

a=-1m/s

• c)total distance covered will be the sum of distances in the first, second and third region

in the first region,

S(1)

$= ut + \frac{1}{2} a {t}^{2} \\ = 0 \times t + \frac{1}{2}× 2 ×{10}^{2} \\ = 100m \\ = 0.1km$

in the second region,

S(2)

= u×t. (since, acceleration is zero)

=20×30×60

=36,000 m

=36km

in the third region,

S(3)

$= ut + \frac{1}{2} a {t}^{2} \\ = 20 \times 20 - \frac{1}{2} \times 1 \times {20}^{2} \\ = 400 - 200 \\ = 200m \\ = .2km$

therefore, total distance,

S=s(1)+s(2)+s(3)

=0.1+36+0.2

=36.3 km

hope it helps you... thank you...!!