the speed of train is reduced from 60km/h at the same time as it travel a distance of 450m if retardation is uniform find how much further it will travel approx befor coming rest?
Since the speed is reducing,it means it's slowing down.the u>v,(v^2-u^2/2as) and apply that...thank you
To find out how much further the train will travel before coming to rest, we need to use the equations of motion for uniformly retardation. The equation of motion we will use is:
v^2 = u^2 - 2as,
where:
v = final velocity (0 m/s, as the train comes to rest)
u = initial velocity (60 km/h converted to m/s)
a = acceleration (uniform retardation)
s = distance traveled
First, let's convert the initial velocity from km/h to m/s:
u = 60 km/h * (1000 m/1 km) * (1 h/3600 s)
u = 16.67 m/s (approx.)
Now we can plug the values into the equation:
0^2 = (16.67)^2 - 2a(450)
Simplifying the equation:
0 = 277.56 - 900a
Rearranging the equation:
900a = 277.56
Dividing both sides by 900:
a = 0.3084 m/s^2 (approx.)
Now, to find out how much further the train will travel before coming to rest, we'll use the equation:
v^2 = u^2 - 2as,
Since the final velocity (v) is 0 m/s, the equation becomes:
0 = (16.67)^2 - 2(0.3084)s
Simplifying the equation further:
0 = 277.56 - 0.6168s
Rearranging the equation:
0.6168s = 277.56
Dividing both sides by 0.6168:
s ≈ 450.16 m
Therefore, the train will travel approximately 450.16 meters further before coming to rest.