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.
Answers
Answered by
1722
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.
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.
Answered by
963
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
Distance = 3 km Ans.
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
Distance = 3 km Ans.
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