How Long Does It Take to Brake a Train from 28.6 m/s to 11.4 m/s?

[lockmantican]

A train approaches a small town with a constant velocity of +28.6 m/s. The operator applies the brake, reducing the train’s velocity to +11.4 m/s. If the average acceleration of the train during braking is –1.35 m/s^2, for what elapsed time does the operator apply the brake?



2. i tried to use the equation for definition of acceleration



3. i set up my diagram and my question is; what is the initial velocity when the operator starts applying the brakes?

 

[aftershock]

A train approaches a small town with a constant velocity of +28.6 m/s. The operator applies the brake, reducing the train’s velocity to +11.4 m/s. If the average acceleration of the train during braking is –1.35 m/s^2, for what elapsed time does the operator apply the brake?



2. i tried to use the equation for definition of acceleration



3. i set up my diagram and my question is; what is the initial velocity when the operator starts applying the brakes?

The train is traveling at 28.6m/s when it starts to experience a negative acceleration of 1.35m/s^2 until it is traveling at 11.4m/s

So initial velocity is 28.6, final velocity is 11.4 and acceleration is -1.35

Just look at your kinematics equations, you should be able to use

Vf = Vo +at

 

[lockmantican]

The train is traveling at 28.6m/s when it starts to experience a negative acceleration of 1.35m/s^2 until it is traveling at 11.4m/s

So initial velocity is 28.6, final velocity is 11.4 and acceleration is -1.35

Just look at your kinematics equations, you should be able to use

Vf = Vo +at

why wouldn't the final velocity be 0 m/s

 

[aftershock]

why wouldn't the final velocity be 0 m/s

Why would it be? The train is starting at 28.6m/s and ending up at 11.4m/s so final velocity is 11.4

The final velocity would be zero if the train came to a stop, but it doesn't.

 

[asteriatic]

The initial velocity when the operator starts applying the brakes is +28.6 m/s. This is the velocity at which the train was traveling before the brakes were applied. The change in velocity, or the deceleration, is 28.6 m/s - 11.4 m/s = 17.2 m/s. This change in velocity occurs over a period of time, which can be calculated using the average acceleration and the formula for acceleration: a = (v - u)/t, where a is acceleration, v is final velocity, u is initial velocity, and t is time. Rearranging this formula, we get t = (v - u)/a. Plugging in the values we know, we get t = (11.4 m/s - 28.6 m/s)/-1.35 m/s^2 = 17.2 m/s / 1.35 m/s^2 = 12.74 seconds. This means that the operator applied the brakes for 12.74 seconds to decelerate the train from +28.6 m/s to +11.4 m/s.