. A train is traveling down a straight track at 20 m/s when the engineer applies the brakes,
resulting in an acceleration of -1.0 m/s
2
as long as the train is in motion. How far does the
train move during a 40-s time interval starting at the instant the brakes is applied?
v =Vi + a t = 20 -1.0 t
the train moves forward until v = 0, then backward
How far it moves includes both directions
Lets do eastward
0 = 20 - 1.0 t
t = 20 seconds to stop
x = Xi + Vi t - (1/2) a t^2
x = 0 + 20(20) - 0.5 (20)^2
x = 200 meters
now it does backwards, you can just say back 200 meters due to the symmetry
so 200 forward and 200 backward = 400 meters total
by the way the average speed during each phase is (20+0)/2 = 10 m/s
so you could have just said 10 * 20 = 200 meters each way
Well, well, well, it looks like we have a braking train on our hands. Let's see how far this train manages to move during its little adventure.
We know the initial velocity of the train is 20 m/s, and the acceleration is a negative 1.0 m/s^2. Now, we have a 40-second time interval to work with. Using the kinematic equation:
distance = initial velocity * time + (1/2) * acceleration * time^2
Substituting the given values:
distance = 20 m/s * 40 s + (1/2) * (-1.0 m/s^2) * (40 s)^2
Calculating this lovely equation, we get:
distance = 800 m - 800 m
Uh-oh! Looks like the train didn't move an inch during this time interval because the negative acceleration exactly counteracted its initial velocity. It's a classic example of "The train that didn't want to budge."
I hope you enjoyed this little anticlimactic train ride. If you have any more questions, feel free to ask!
To find the distance the train moves during a 40-second time interval, we can use the kinematic equation:
d = vit + (1/2)at^2
where:
d = distance
vi = initial velocity
a = acceleration
t = time
Given:
vi = 20 m/s
a = -1.0 m/s^2 (negative because it's deceleration)
t = 40 s
Substituting the given values into the equation:
d = (20 m/s)(40 s) + (1/2)(-1.0 m/s^2)(40 s)^2
Simplifying:
d = 800 m - (1/2)(-1.0 m/s^2)(1600 s^2)
d = 800 m - (-800 m)
d = 800 m + 800 m
d = 1600 m
Therefore, the train moves a distance of 1600 meters during the 40-second time interval.
To find the distance the train moves during the time interval, we can use the equations of motion. The equation that relates distance (d), initial velocity (u), acceleration (a), and time (t) is:
d = ut + (1/2) * a * t^2
Given:
Initial velocity (u) = 20 m/s
Acceleration (a) = -1.0 m/s^2 (negative sign indicates deceleration)
Time interval (t) = 40 s
Substituting the given values into the equation, we have:
d = (20 m/s) * (40 s) + (1/2) * (-1.0 m/s^2) * (40 s)^2
Simplifying the equation:
d = 800 m - 800 m
The term (1/2) * (-1.0 m/s^2) * (40 s)^2 evaluates to -800 m, which cancels out the positive 800 m, resulting in a distance of 0 m.
Therefore, during the 40-s time interval starting at the instant the brakes are applied, the train does not move any further and remains at the same position.