A train starting from rest attains a velocity of 72 kmh–1 in 5 minutes...
Solution:
Given:
Initial velocity, u = 0 kmh–1
Final velocity, v = 72 kmh–1
Time taken, t = 5 min = 5/60 h
To find:
(i) Acceleration, a
(ii) Distance travelled, s
Formula used:
We know that,
v = u + at
s = ut + 1/2 at^2
Where,
v = Final velocity
u = Initial velocity
a = Acceleration
t = Time taken
s = Distance travelled
Calculation:
(i) Acceleration, a
We know that,
v = u + at
a = (v - u) / t
Substituting the given values, we get
a = (72 - 0) / (5/60)
a = 86.4 kmh–2
Therefore, the acceleration of the train is 86.4 kmh–2.
(ii) Distance travelled, s
We know that,
s = ut + 1/2 at^2
Substituting the given values, we get
s = 0 x (5/60) + 1/2 x 86.4 x (5/60)^2
s = 0.3 km
Therefore, the distance travelled by the train to attain the velocity of 72 kmh–1 is 0.3 km.
Conclusion:
Thus, the acceleration of the train is 86.4 kmh–2 and the distance travelled by the train to attain the velocity of 72 kmh–1 is 0.3 km.
A train starting from rest attains a velocity of 72 kmh–1 in 5 minutes...
Here, u=0. v= 72km/h = 72× 5/18 = 20 m/s t= 5min = 300s
a= v- u / t 20- 0/ 300= 20 / 300 = 0.067m/s
^2
s= ut + 1/2 at^2. s= 0× t + 1/2 ×0.067×300×300. s = 1005m
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