A train traveling at 72km/h undergoes at uniform retardation of -2m/s when brake were applied.find the time taken to come rest and the distance travel on the place where brake were apply.
you want to find t when v=0
72 km/hr * 1000m/km * 1hr/3600s = 20 m/s
v = 20 - 2t
s = 20t - t^2
now find t and s
To find the time taken for the train to come to rest and the distance traveled when the brakes were applied, we can use the equations of motion.
First, convert the speed of the train from km/h to m/s:
72 km/h = 20 m/s (since 1 km/h = 1000 m/3600 s = 10/36 m/s)
Given:
Initial speed (u) = 20 m/s
Retardation (a) = -2 m/s² (since retardation is negative when the brakes are applied)
We can use the equation of motion:
v = u + at
where:
v = final velocity (0 m/s when the train comes to rest)
u = initial velocity
a = acceleration (negative retardation in this case)
t = time taken
Substituting the values into the equation, we have:
0 = 20 + (-2)t
Simplifying the equation:
-2t = -20
Dividing by -2:
t = 10 seconds
So, it will take 10 seconds for the train to come to rest.
To find the distance traveled when the brakes were applied, we can use the equation:
s = ut + (1/2)at²
where:
s = distance traveled
u = initial velocity
t = time taken
a = acceleration (negative retardation in this case)
Substituting the values into the equation, we have:
s = 20(10) + (1/2)(-2)(10)²
Simplifying:
s = 200 - 100
s = 100 meters
Therefore, the distance traveled on the place where the brakes were applied is 100 meters.
To find the time taken to come to rest and the distance traveled where the brakes were applied, we need to use the equations of motion for uniform acceleration/deceleration.
Let's break down the problem step-by-step:
1. Convert the speed of the train from km/h to m/s:
We know that 1 km/h is equal to 1000/3600 m/s. So, the speed of the train is:
72 km/h = (72 x 1000) / 3600 m/s = 20 m/s
2. Determine the acceleration (retardation) of the train:
The given value of retardation is -2 m/s². Negative sign indicates retardation (deceleration).
3. Find the time taken to come to rest:
We can use the equation of motion: v(final) = v(initial) + at,
where v(final) is the final velocity (0 m/s), v(initial) is the initial velocity (20 m/s), a is the acceleration (retardation), and t is the time taken.
Plugging the values into the equation, we have:
0 = 20 + (-2)t
Solving for t:
2t = 20
t = 20/2
t = 10 seconds
Therefore, the time taken to come to rest is 10 seconds.
4. Determine the distance traveled where the brakes were applied:
We can use the equation of motion: s = ut + (1/2)at²,
where s is the distance traveled, u is the initial velocity, t is the time taken, and a is the acceleration (retardation).
Plugging the values into the equation, we have:
s = (20 x 10) + (1/2)(-2)(10²)
Simplifying the equation:
s = 200 + (-1)(100)
s = 200 - 100
s = 100 meters
Therefore, the distance traveled where the brakes were applied is 100 meters.