Question

train starts moving from left and acquire velocity of 72 km per hour in 5 minutes if the train move at constant acceleration calculate the following first distance travelled by the train to acquire velocity

Answer 1

Given :
Initial Velocity=u=0m/s
Final velocity= V= 72km/h
Time =5 min =5/60 h =1/12 h
acceleration=a=?
As we know that , a=v-u/t
⇒a=72-0/1/12
=72*12
=864km/h²
Distance travelled  =s=ut+1/2at²
=0x1/12 +1/2(864)(1/12)²
=(1/2)x864x(1/144)
=3km
∴Distance travelled by the train for attaining this velocity 3km.

THANKS!!



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.

Given :
Initial Velocity=u=0m/s
Final velocity= V= 72km/h
Time =5 min =5/60 h =1/12 h
acceleration=a=?
As we know that , a=v-u/t
⇒a=72-0/1/12
=72*12
=864km/h²
Distance travelled  =s=ut+1/2at²
=0x1/12 +1/2(864)(1/12)²
=(1/2)x864x(1/144)
=3km
∴Distance travelled by the train for attaining this velocity 3km.

THANKS

Given,

Initial velocity u = 0

Final velocity v = 72 km/hr

Time taken t = 5 min

= 5/60 hr

= 1/12 hr

Acceleration a = ?

Distance travelled s = ?

Calculation,

i) Acceleration:-

We know that, a = (v - u) / t

a = (72 - 0) / (1/12)

a = 72 / (1/12)

a = (72 * 12) / 1

a = 72 * 12

a = 864 km/hr²

Acceleration = 864 km/hr² Ans. 

ii) The distance travelled by the train for attaining this velocity:-

s = ut + 1/2at² (Second equation of motion)

s = 0 * (1/12) + 1/2 * 864 * (1/12)²

s = 1 * 432 * (1/144)

s = 432 / 144

s = 3 km

Now,1km=1000m
So,3km=3000m

Distance = 3 km Ans.

Answer 2

5/18 is the conversion factor for km/hr to m/s