Question

A train starting from rest attains a velocity of 20m/s in one minute. Assuming that the acceleration is uniform, find a) the acceleration, b) the distance travelled by the train to attain the velocity

Answer 1

Answer:

Explanation:

Solution,

Here, we have

Initial velocity, u = 0 (As train starts from rest)

Final velocity, v = 20 m/s

Time taken, t = 1 min = 60 seconds

To Find,

a. Acceleration, a = ?

b. Distance covered, s = ?

According to the 1st equation of motion,

We know that

v = u + at

So, putting all the values, we get

⇒ v = u + at

⇒ 20 = 0 + a × 60

⇒ 20 = 60

⇒ 20/60 = a

⇒ a = 0.33 m/s²

Hence, the acceleration of train is 0.33 m/s².

Now, the distance covered, s,

According to the 3rd equation of motion,

We know that

v² - u² = 2as

So, putting all the values again, we get

⇒ v² - u² = 2as

⇒ (20) - (0) = 2 × 0.33 × s

⇒ 400 = 0.66s

⇒ 400/0.66 = s

⇒ s = 606.06 m.

Hence, the distance covered by train is 606.06 m.

Answer 2

Given:-

• Initial velocity (u) of train =0 m/s

• Final velocity (v) of train=20m/s

• Time taken (t) = 1min = 1×60= 60s

To Find:-

• (a) The acceleration (a) of train.

• (b) The distance (s) covered by train.

Solution :-

By using 1st equation of motion

➦v = u +at

➭ 20 = 0 + a × 60

➭ 20 = 60×a

➭ a =20/60

➭ a = 1/3 m/s²

∴The acceleration of train is 1/3 m/s²

And, Now in Second case

we have to find the distance covered by train to attain this velocity.

So, by using 2nd equation of motion

➦s = ut +1/2 at²

➭ s = 0×60+1/2 × 1/3 × 60²

➭ s = 1/2 ×1/3 × 3600

➭ s = 1/3 × 1800

➭ s = 600m

∴ The distance covered by train to attain the velocity is 600m.