Question

A train starting from rest attain the speed of 108 km in 2 minutes , calculate its acceleration and distance covered by it​

Answer 1

Answer:

• Acceleration = 0.25m/s²
• Distance covered = 3600m

Explanation:

Given that,

A train starting from rest attains a speed of 108km/h in 2 minutes

Converting 108km/h into m/s :-

= 108 ÷ 3.6

= 30m/s

Now, Acceleration is given by :-

$\sf \longrightarrow \: a = \dfrac{(v - u)}{t}$

Where,

• v(final velocity) = 30m/s
• u(initial velocity) = 0m/s [Since, it at rest]
• t(time) = 2 minutes or, 120 seconds.

From the given parameters,

$\sf \longrightarrow \: a = \dfrac{(30 - 0)}{120}$

$\sf \longrightarrow \: a = \dfrac{30}{120}$

$\sf \longrightarrow \: a = 0.25 {m/s}^{2}$

∴ Acceleration of the train is 0.25m/s²

And also, calculating the distance :-

→ Distance = Speed × Time

→ 30 × 120

→ 3600m

∴ Distance covered is 3600m

______________________

Formula used:

$\sf \bullet \: a = \dfrac{(v - u)}{t}$

Where,

• v = final velocity
• u = initial velocity
• t = time taken

Distance = Speed × time

Conversion:

To change km/h into m/s we divided the speed by 3.6