A braking train can decelerate at 0.4m/s^2. If the train is traveling at 160km/hr and needs to plan to stop at a pick-up area. How far in advance does it need to apply the brakes?
I know I need to convert the 160km/hr into m/s, which I did, it's 44.4m/s, but I'm super confused on how to solve it from there.
Thanks for the help!
To solve this problem, you can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, since the train needs to stop)
u = initial velocity (44.4 m/s)
a = acceleration/deceleration (-0.4 m/s^2)
s = distance
Rearranging the equation to solve for distance (s), we have:
s = (v^2 - u^2) / (2a)
Now, let's substitute the given values into the equation:
s = (0^2 - 44.4^2) / (2 * -0.4)
Calculating this, we get:
s = 19782 m
Hence, the train needs to start applying the brakes at least 19782 meters (or approximately 19.8 kilometers) in advance to come to a complete stop at the pick-up area.
To solve this problem, you need to use the kinematic equation that relates distance, initial velocity, final velocity, and acceleration:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, since the train needs to stop)
u = initial velocity (44.4 m/s, the converted speed of the train)
a = acceleration (0.4 m/s^2, the deceleration of the train)
s = unknown distance to be solved
You can rearrange the equation to solve for s:
s = (v^2 - u^2) / (2a)
Plugging in the values, you get:
s = (0 - 44.4^2) / (2(-0.4))
s = (0 - 1971.36) / (-0.8)
s = 2464.2 meters
So the train needs to start applying the brakes 2464.2 meters in advance in order to stop at the pickup area.
divide the velocity by the deceleration to find the stopping time
(44.4 m/s) / (0.4 m/s^2) = ? s