A train is travelling at a speed of 90 km h^-1
. Brakes are applied so as to produce a uniform acceleration of −0.5 ms ^-2
. Find how far the train will go before it is brought to rest?
Answers
A train travelling with a speed of 90 km/hr. Here, the initial velocity/speed of the train is 90 km/hr.
Later, brakes are applied so the produce uniform of -0.5 m/s². (Final velocity of the train is 0 m/s as brakes are applied and acceleration of the train is -0.5 m/s², negative sign shows retardation).
We have to find the distance travelled by the train before it brought to rest.
First equation of motion:
v = u + at
Second equation of motion:
s = ut + 1/2 at²
Third equation of motion:
v² - u² = 2as
From above data we have, v = 0 m/s, u = 90 × 5/18 = 25 m/s and a = -0.5 m/s².
So, using the Third Equation Of Motion,
v² - u² = 2as
Substitute the known in the above formula,
→ (0)² - (25)² = 2(-0.5)(s)
→ 0 - 625 = -1s
→ 625 = 1s
→ s = 625
Therefore, the distance travelled by the train is 625 m before it brought to rest.
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