A train is traveling at a speed of 75km/hr. The brakes are applied to produce a uniform acceleration of -0.5 m/s2.Find how far the train goes before it stops? .
Answers
Given :-
- initial velocity of train, u = 75 km/h
- acceleration produced on applying brakes, a = -0.5 m/s²
To find :-
- Distance traveled by train after applying brakes and before coming to rest.
Formula used :-
Third eqn of motion
2 as = v² - u²
(where a is the acceleration, s is the distance traveled, u is the initial velocity and v is the final velocity)
Solution :-
Converting initial velocity given in Km/h into m/s
u = 75 km /h
u = 75 × 5 / 18
▶ u = 20.84 m/s
Since, Train comes to rest hence,
final velocity of train, v = 0 m/s
and Let, distance traveled by train = s
Using third eqn of motion
➠ 2 a s = v² - u²
putting known values
➠ 2 (-0.5) s = (0)² - (20.84)²
➠ - s = - 434.3
➠ s = 434.3 m
Hence,
Train will travel a distance of 434.3 metres. (approximately)
More equations of motion :-
● First eqn of motion
v = u + at
● Second eqn of motion
s = ut + 1/2 at²
(where u is the initial velocity, v is final velocity, a is the acceleration , t is time taken and s is the distance travelled by object).