An express subway train passes through an underground station. It enters at t = 0 with an initial velocity of 23.0 m/s and decelerates at a rate of 0.150 m/s2 as it goes through. The station in 210 m long.

(a) How long is the nose of the train in the station?

(b) How fast is it going when the nose leaves the station?

(c) If the train is 130 m long, at what time t does the end of the train leave the station?

(d) What is the velocity of the end of the train as it leaves?

Additional Materials

(a) use d=vi*t+0.5*a*t^2

manipulate the equation to solve for t

210 m = (23 m/s)t+0.5(0.150 m/s^2)*t^2
t = 9.41 s

(b) v = d/t
= 210 m/ 9.41 s
= 22.3 m/s

I'm still trying to figure out the rest...

3.60

To solve this problem, we can use the equations of motion for linear motion.

Given:
Initial velocity, u = 23.0 m/s
Deceleration, a = -0.150 m/s^2 (negative because it is decelerating)
Length of the station, s = 210 m
Length of the train, l = 130 m

(a) To find how long the nose of the train is in the station, we need to calculate the time it takes for the train to travel the length of the station.

We can use the equation:

s = ut + (1/2)at^2

where:
s is the displacement (210 m),
u is the initial velocity (23.0 m/s),
t is the time taken,
and a is the deceleration (-0.150 m/s^2).

Plugging in the given values, we have:

210 = 23.0t + (1/2)(-0.150)t^2

Rearranging the equation, we get a quadratic equation:

0.075t^2 + 23.0t - 210 = 0

Now we can solve this quadratic equation to find the values of t. Using the quadratic formula, we have:

t = (-b ± √(b^2 - 4ac))/2a

Substituting the values into the formula, we have:

t = (-23.0 ± √(23.0^2 - 4(0.075)(-210)))/(2(0.075))

Simplifying this equation will give us two possible values for t. We will consider the positive value since time cannot be negative.

(b) To find the speed of the train when the nose leaves the station, we need to calculate the final velocity, v.

We can use the equation:

v = u + at

where:
v is the final velocity,
u is the initial velocity (23.0 m/s),
a is the acceleration (-0.150 m/s^2),
and t is the time taken.

Plugging in the given values, we have:

v = 23.0 + (-0.150)t

(c) To find the time when the end of the train leaves the station, we need to calculate the time it takes for the end of the train (130 m) to travel the length of the station (210 m).

Using the equation s = ut + (1/2)at^2, we have:

210 = 23.0t + (1/2)(-0.150)t^2

Simplifying this equation will give us another quadratic equation. We can solve this equation using the quadratic formula to find the value of t.

(d) To find the velocity of the end of the train as it leaves the station, we can use the same equation as in part (b).

v = 23.0 + (-0.150)t

To solve these problems, we can use the equations of motion to find the answers step by step. Here's how to approach each part of the question:

(a) To find the length of the nose of the train in the station, we need to find the time it takes for the train to come to a stop. Since we know the initial velocity, the deceleration, and the distance traveled, we can use the equation:

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s)
u = initial velocity (23 m/s)
a = acceleration (-0.150 m/s^2)
s = distance (210 m + length of the nose)

Rearranging the equation, we get:

s = (v^2 - u^2)/(2a)

Plugging in the values, we get:

s = (0^2 - (23)^2)/(2*(-0.150))

Simplifying, we get:

s = -529/(-0.30) = 1763.33 m

The length of the nose of the train is the total length of the train minus the station length:

Length of the nose = 1763.33 m - 210 m = 1553.33 m (rounded to 2 decimal places)

(b) To find the speed of the train when the nose leaves the station, we use the equation:

v = u + at

Where:
v = final velocity (unknown)
u = initial velocity (23 m/s)
a = acceleration (-0.150 m/s^2)
t = time

We know the station length, so we can set the distance traveled equal to the station length to find the time it takes for the train to stop. Using the equation:

s = ut + (1/2)at^2

Where:
s = distance (210 m)
u = initial velocity (23 m/s)
a = acceleration (-0.150 m/s^2)
t = time

Plug in the values and solve for t:

210 = 23t + (1/2)(-0.150)t^2

Rearranging the equation, we get a quadratic equation:

-0.075t^2 + 23t - 210 = 0

Solving this equation, we find two possible values for t: t = 4.275 s and t = -5.725 s. Since time cannot be negative, we disregard the negative value.

So, the train takes 4.275 seconds to come to a stop.

To find the speed when the nose leaves the station, plug in the values into the equation:

v = u + at

v = 23 + (-0.150)(4.275)
v = 23 - 0.641
v = 22.359 m/s (rounded to 3 decimal places)

The speed of the train when the nose leaves the station is approximately 22.359 m/s.

(c) To find the time when the end of the train leaves the station, we need to find the total time it takes for the train to pass through the station. Since we already know the time it takes for the train to stop (4.275 s), we can use this time to find the time it takes for the end of the train to pass the station.

The total time is given by:

Total time = time to stop + time for 130 m to pass

Total time = 4.275 s + (130 m/23 m/s)
Total time = 4.275 s + 5.652 s
Total time = 9.927 s (rounded to 3 decimal places)

So, the end of the train leaves the station at approximately 9.927 seconds.

(d) To find the velocity of the end of the train as it leaves the station, we can use the equation:

v = u + at

Where:
v = final velocity (unknown)
u = initial velocity (23 m/s)
a = acceleration (-0.150 m/s^2)
t = time (9.927 s)

Plug in the values and solve for v:

v = 23 + (-0.150)(9.927)
v = 23 - 1.489
v = 21.511 m/s (rounded to 3 decimal places)

So, the velocity of the end of the train as it leaves the station is approximately 21.511 m/s.