A train travelling at 20m/s undergoes a uniform retardation of 2m/s(square) when brakes are applied calculate the time taken to come to rest and the distance travelled from the place where brakes where applied
v = v0 + at = 20 - 2t
so, when is v=0?
now recall that
s = v0*t + 1/2 at^2 = 20t - t^2
plug in your value for t
please continue with the solution
A train travell at 20ms-1 undergone a umform retardation of 2m-2 when breakers are applied calculate the time taken to come to rest and the distance travelled from the place where the breakers were applied
To calculate the time taken for the train to come to rest and the distance traveled from the place where the brakes were applied, we can use the equations of motion.
First, let's identify the known values:
Initial velocity (u) = 20 m/s
Acceleration (a) = -2 m/s² (negative because the train is slowing down)
Final velocity (v) = 0 m/s (since the train comes to rest)
Distance traveled (s) = ?
We can use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken. Rearranging the equation, we have t = (v - u) / a.
Substituting the known values, we get:
t = (0 - 20) / -2
t = 20 / 2
t = 10 seconds
Therefore, it takes 10 seconds for the train to come to rest.
To calculate the distance traveled (s), we can use the equation: s = ut + (1/2)at². Substituting the known values:
s = (20 * 10) + (1/2)(-2)(10²)
s = 200 + (-1)(100)
s = 200 - 100
s = 100 meters
Therefore, the train travels a distance of 100 meters from the place where the brakes were applied.