Question

A train is moving with a velocity of 18 km/hr to 36 km/hr in 5 mins in a

uniform velocity. Find its acceleration and distance covered by the train.​

Answer 1

18 km/hr = 18(1000/3600) m/s = 5m/s

36 km/hr = 10 m/s

time = 5 min = 300 sec

Acceleration = Δv/Δt

            = (10 - 5)/(300 - 0)

            = 5/300

            = 0.0167 m/s²

Using, v² = u² + 2aS

⇒ (10)² = 5² + 2(5/300)S

⇒ 100 - 25 = (1/30)S

⇒  75(30) = S

⇒ 2250= S      

Or, 2.250 km

  ∴ Acceleration = 0.0167 m/s²

      Distance = 2.25 km

Note that: If t = 5 sec* , answers are 1m/s²  &  37.5 m  following the same procedure  [if not, ignore this]

Answer 2

Given :-

Initial velocity = 18 km/h

Final velocity = 36 km/h

Time = 5 min

To Find :-

Acceleration

Distance covered

Solution :-

First we need to change the unit

36 × 5/18 = 2 × 5 = 10 m/s

18 × 5/18 = 5 m/s

5 min = 300 sec

v = u + at

10 = 5 + a(300)

10 - 5 = a(300)

5 = 300a

5/300 = a

0.0167 = a

Finding distance covered

${\mathfrak{\pink{v^{2} - u^{2} = 2as}}}$

$\sf 10^2 - 5^2 = 2(0.0167)(s)$

$\sf 100 - 25 = 0.0334s$

$\sf 75 = 0.0334s$

$\sf s = \dfrac{75}{0.0334}$

$\sf s = 2250 \;m$