Question

A train starting from rest attains a velocity of 90 km/h in 2 minutes. Assuming the acceleration to be uniform, find the value of a) acceleration b) distance travelled in 2 minutes​

Answer 1

Answer:

a) 0.21 m/s²

b) 1488 metres

Explanation:

Given :

• Initial velocity = u = 0 m/s

• Final velocity = v = 90 km/h

• Time taken = t = 2 minutes

To find :

• Acceleration of the train

• Distance travelled by the train in 2 minutes

90 km/h = 90×5/18 = 25 m/s

Time taken = 2 minutes = 120 seconds

Using the first equation of motion :

V=u+at

25=0+a×120

25=120a

a = 25/120 = 5/24 = 0.21 m/s²

Now using the third equation of motion :

V²-u²=2as

25²-0²=2×0.21×s

625=0.42×s

s = 625/0.42

S ≈ 1488 metres (approx)

The acceleration of the train is 0.21 m/s² and distance travelled is equal to 1488 metres approximately

Answer 2

Answer :

Given that a train starting from rest attains a velocity of 90 km/h in 2 minutes.

We have to find a) Acceleration b) Distance travelled in 2 minutes.

From the data we have :

Initial velocity (u) = 0 m/s

Final velocity (v) = 90 km/h = 90 × (5/18) = 25 m/s

Time (t) = 2 minutes. = (2 × 60) = 120 s

From the first equation of motion :

→ v = u + at

Substituting values,

→ 25 = 0 + a × 120

→ 25 = 120a

→ a = 25/120

→ a = 0.21 m/s²

∴ Acceleration of train = 0.21 m/s²

Now using 2nd equation of motion :

→ s = ut + ½ at²

Substituting values,

→ s = 0 × 120 + ½ × 0.21 × (120)²

→ s = 0 + ½ × 0.21 × 14400

→ s = 0.21 × 7200

→ s = 1512 m

∴ Distance travelled in 2 minutes = 1512 m