Question

A train starts from rest and moves with a constant
acceleration for the first 1 km; for the next 3km, it
has a constant velocity and for another 2km, it
moves with constant retardation to come to rest
after 10 min. Find the maximum velocity and the
time intervals in the three parts of motion.​

Answer 1

Answer:

That's your answer I hope it helps you

Answer 2

Maximum Velocity = $\frac{9}{10}$ km/min

Time taken in $1^{st}$ Part = $\frac{20}{9}$ min

Time taken in $2^{nd}$ Part = $\frac{30}{9}$min = $\frac{10}{3}$ min

Time taken in $3^{rd}$ Part = $\frac{40}{9}$ min

Explanation:

Part 1: Constant Acceleration

Let the constant acceleration for the 1st 1km be 'a' units.

s = ut + $\frac{at^{2} }{2}$ (∵ u =0 and s = 1km)

1 =   $\frac{at^{2} }{2}$  ⇒ t = $\sqrt{\frac{2}{a} }$ min

∴ $t_{1}$ =  $\frac{2}{\sqrt{2a} }$ min

$v^{2}$ - $u^{2}$ = 2as

∴ v = $\sqrt{2a}$ km/min

Part 2: Constant Velocity

s = vt ( ∵ s = 3km and v = $\sqrt{2a}$ km/min )

∴ $t_{2}$ = $\frac{3}{\sqrt{2a} }$ min

Part 3: Constant Retardation

Let the magnitude of constant retardation be 'b' units

∴ Acceleration = -b km/$min^{2}$

s = vt - $\frac{at^{2} }{2}$  ( ∵ s = 2km , v = 0 and a = -b)

2 = $\frac{bt^{2} }{2}$  ⇒ t = $\frac{2}{\sqrt{b} }$ min

$v^{2}$ - $u^{2}$ = 2as ( ∵ s = 2km , u =  $\sqrt{2a}$ km/min , v = 0 and a = -b)

-2a = 2(-b)×2 ⇒ b = $\frac{a}{2}$

Substitute b = $\frac{a}{2}$ in  t = $\frac{2}{\sqrt{b} }$

t = $\frac{2\sqrt{2} }{\sqrt{a} }$ min

$t_{3}$ = $\frac{4}{\sqrt{2a} }$ min

Given that total time of motion is 10 min.

∴ $t_{1}$ +  $t_{2}$ + $t_{3}$ = 10 min

$\frac{2}{\sqrt{2a} }$ +  $\frac{3}{\sqrt{2a} }$ +  $\frac{4}{\sqrt{2a} }$  = 10

∴ $\frac{9}{\sqrt{2a} }$  = 10

∴  $\sqrt{2a}$  =  $\frac{9}{10}$

Therefore,

Maximum Velocity = $\frac{9}{10}$ km/min

Time taken in $1^{st}$ Part = $\frac{20}{9}$ min

Time taken in $2^{nd}$ Part = $\frac{30}{9}$min = $\frac{10}{3}$ min

Time taken in $3^{rd}$ Part = $\frac{40}{9}$ min