Question

A train starting from rest attains a velocity of 108 km/h in 5 minutes. Assuming that
the acceleration is uniform, find out the
(i) acceleration
(ii) the distance travelled by the train for attending this velocity?

Answer 1

Answer:

(i) 0.10 m/s²

(ii)4500m

Explanation:

(i) v=108×5/18, t=5×60

v=30m/s , t=300s

v=u+at

30=0+a×300

30/300=a

a=0.10m/s²

(ii)s=ut+1/2at²

s=0×300+1/2×0.10×(300)²

s=1/20×90000

s=4500

Answer 2

• (i) The acceleration of the train = 0.1 m/s²

• (ii) The distance travelled by the train for attending this velocity = 4500 m

Given :

• A train starting from rest attains a velocity of 108 km/h in 5 minutes

• Assume that the acceleration is uniform

To find :

(i) The acceleration of the train

(ii) The distance travelled by the train for attending this velocity

Formula Used :

First equation of motion

$\displaystyle \sf v = u + at$

Second equation of motion

$\displaystyle \sf S = ut + \frac{1}{2} a {t}^{2}$

Solution :

Step 1 of 3 :

Write down the given data

Initial velocity = u = 0

$\displaystyle \sf Final \: velocity = v = 108 \: km/h = \frac{108 \times 1000}{3600} \: m/s = 30\: m/s$

Time

= t

= 5 minutes

= 5 × 60 seconds

= 300 seconds

Step 2 of 3 :

Calculate acceleration of the train

Let acceleration of the train = a m/s

From First equation of motion

$\displaystyle \sf v = u + at$

$\displaystyle \sf{ \implies }30 = 0 + (a \times 300)$

$\displaystyle \sf{ \implies }300a = 30$

$\displaystyle \sf{ \implies }a = \frac{30}{300}$

$\displaystyle \sf{ \implies }a = \frac{1}{10}$

$\displaystyle \sf{ \implies }a = 0.1$

The acceleration of the train = 0.1 m/s²

Step 3 of 3 :

Calculate distance travelled by the train for attending this velocity

Let distance travelled by the train for attending this velocity = S metre

From Second equation of motion

$\displaystyle \sf S = ut + \frac{1}{2} a {t}^{2}$

$\displaystyle \sf{ \implies }S = (0 \times 300) + \frac{1}{2} \times 0.1 \times {(300)}^{2}$

$\displaystyle \sf{ \implies }S = 0 + \frac{1}{2} \times 0.1 \times 90000$

$\displaystyle \sf{ \implies }S = 0.1 \times 45000$

$\displaystyle \sf{ \implies }S = 4500$

∴ The distance travelled by the train for attending this velocity = 4500 m

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