A train is going at 23.1 m/s when the train engineer sees a car on the tracks and pulls the emergency brakes. If it takes the train 25.5 s to fully stop, how far did the train go before it stopped.

Answer 1

There is more than one way to work this type of problem. Michael gave you a good solution. Here is another way:

d = distance

v₀ = initial speed = 23.1 m/s

v = final speed = 0 m/s (full stop)

t = time interval to stop = 25.5 s

a = acceleration

d = (1/2)at² + v₀t

a = (v - v₀)/t = [(0 - 23.1)/(25.5)] m/s² = ?

v₀ = initial speed = 23.1 m/s

v = final speed = 0 m/s (full stop)

t = time interval to stop = 25.5 s

a = acceleration

d = (1/2)at² + v₀t

a = (v - v₀)/t = [(0 - 23.1)/(25.5)] m/s² ≅ -0.9059 m/s²

Note that a is negative.

d = [-0.9059(25.5)²/2 + 23.1(25.5)] m ≅ 295 m

Answer 2

If we assume constant deceleration (negative acceleration), then we can compute the average velocity as the train slows down from full speed (23.1 m/s) to a full stop (0 m/s).

V(ave) = (23.1 m/s - 0 m/s) / 2 = 11.55 m/s

Distance (x) = V(ave) * t = (11.55 m/s) * (25.5 s) = 294.525 m

Therefore, the train traveled 294.5 meters before it stopped.

Answer 3

The train's average speed was (23.1 + 0) / 2 = 23.1/2 m/s.

Over 25.5 sec, it traveled (23.1 / 2) * 25.5 = 294.525 m

Based on the precision of the information provided,

we justified in answering with only 3 significant figures: 295 m.

Answer 4

I will set it up for you.

d = distance

v₀ = initial speed = 23.1 m/s

v = final speed = 0 m/s (full stop)

t = time interval to stop = 25.5 s

a = acceleration

d = (1/2)at² + v₀t

a = (v - v₀)/t = [(0 - 23.1)/(25.5)] m/s² = ? m/s²

Note that the acceleration is negative, and you must use the correct sign when you plug it into the equation for distance. You have all the numbers; just plug them in and get a, then plug a and the other numbers in to get d. You should be able to finish from here.