A train is travelling at a speed of 90km/h. Brakes are applied so as to produce a uniform acceleration of -0.5m/s². Find how far will the train go before it is brought to rest?
Q 2) A bus starting from rest moves with a uniform acceleration of 0.1 m/s² for 2 min Find (a) the speed acquired (b) distance travelled
Answers
Explanation:
1) Initial speed, u = 90 km/h = 90 x (1000/3600) m/s = 25 m/s
final speed, v = 0 m/s
uniform acceleration, a = -0.5 m/s2
Using equation of motion: v2 = u2 + 2as
We have: 0 = (25)2 + 2x(-0.5)xs
s = 625 m
[s: distance traveled by the train after the brakes are applied and before it is finally brought into rest]
2)Acceleration
The bus starts from rest.
Therefore, initial velocity (u) = 0 m/s
Acceleration (a) = 0.1 m.s-2
Time = 2 minutes = 120 s
Acceleration is given by the equation a=(v-u)/t
Therefore, terminal velocity (v) = (at)+u
= (0.1 m.s-2 * 120s) + 0 m.s-1
= 12m.s-1 + 0 m.s-1
Therefore, terminal velocity (v) = 12m/s
(b) As per the third motion equation,
Since a = 0.1 m.s-2, v = 12 m.s-1, u = 0 m.s-1, and t = 120s, the following value for s (distance) can be obtained.
Distance, s =(v2 – u2)/2a
=(122 – 02)/2(0.1)
Therefore, s = 720m.
The speed acquired is 12m.s-1 and the total distance traveled is 720m.