A train starting from rest attains a velocity of 72km/h in 5minutes . Assuming that the acceleration is uniform, - the acceleration , the distance travelled by the train for attaining this velocity
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Answered by
18
Initial velocity (u)=0
Final velocity (v)=72km/h
=20m/s
Time (t)= 5 minute
=5×60 =300s
Acceleration (a)=(v-u)/t
=20/300
=1/15
So, the acceleration is 1/15.
Let distance travelled be s
Using second equation of motion
ut+(1/2)at^2
=(0)(300)+ (1/2)(1/15)(300^2)
= 3000
So the body covered 3000m (3km).
Final velocity (v)=72km/h
=20m/s
Time (t)= 5 minute
=5×60 =300s
Acceleration (a)=(v-u)/t
=20/300
=1/15
So, the acceleration is 1/15.
Let distance travelled be s
Using second equation of motion
ut+(1/2)at^2
=(0)(300)+ (1/2)(1/15)(300^2)
= 3000
So the body covered 3000m (3km).
Answered by
10
Given :
- u = 0
- v = 72 km/h
- t = 5 min
To find :
- the acceleration
- the distance travelled by the train for attaining this velocity
Solution :
u = 0
v = 72 km/h = (72 x 100 m)/(60 x 60 s) = 20 m/s
t = 5 min = 5 x 60 = 300 s
thus,
a = (v - u)/t
= (20 - 0)/300
= 1/15 m/s²
from,
v² - u² = 2 as
s = (v² - u²)/ 2a
= (20² - 0²)/2 x (1 / 15)
= 3000 m
= 3 km
=> 1/15 m/s² is the acceleration and 3 km distance travelled by the train for attaining this velocity
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