Question

Class 9th question A train starting from rest attains a velocity of 72km/hr in 5 min. Assuming that acceleration is uniform. Find (i) acceleration (ii) the distance travelled by the train for attaining this velocity.

Answer 1

v=u+at
a=(v-u) /t
a=(20-0)/300. [convert 720km/hr in m/s]
a=1/15m/s²
s=ut+1/2at²
s=0×300+1/2×1/15×300×300
S=3000m
3km.

Answer 2

✬ Acceleration = 864 km/h² ✬

✬ Distance travelled = 3 km ✬

Explanation:

Given:

• A train stated from rest (u) = 0 m/s
• Train attains final velocity (v) = 72 km/hr.
• Time taken (t) = 5 minutes.

To Find:

• 1.) Acceleration of the train.
• 2.) Distance covered by train.

Formula to be used:

• a = (v – u/t) ( For acceleration )
• s = ut + 1/2at² ( For distance )

Solution: Here, Time is given in minutes so we need to change it into hour. Therefore,

➬ 60 minutes = 1 hours therefore

➬ 5 minutes = 5/60 = 1/12 hour

•Now, Put the values on 1st formula •

$\implies{\rm }$ a = (v – u/t)

$\implies{\rm }$ a = (72 – 0/1/12)

$\implies{\rm }$ a = (72/1/12)

$\implies{\rm }$ a = (72 $\times$ 12)

$\implies{\rm }$ a = 864 km/h²

Hence, the acceleration of the train is 864 km/h².

______________________

• Using second formula for finding distance •

$\implies{\rm }$ s = ut + 1/2at²

$\implies{\rm }$ s = 0 $\times$ 1/12 + 1/2 $\times$ 864 $\times$ (1/12)²

$\implies{\rm }$ s = 0 + 1/2 $\times$ 864 $\times$ 1/144

$\implies{\rm }$ s = 432/144

$\implies{\rm }$ s = 3 km

Hence, the distance travelled by train for attaining final velocity is 3 km.