Train Question? Uniform acceleration

[kencamarador]

A subway trains starts from station A with a uniform acceleration of .5m/s and attains full speed after 50 seconds. It travels at full speed for 120 seconds then is brought to rest at station by by the application of brakes which provides a acceleration of -2.50.

Find the distance between A and B

I divided this solution into 3 stages.

1. Acceleration
2. Constant
3. deacceleration

For the first part, I used Δd=viΔt + 1/2 aav Δt^2

→ 0 x 50 + 1/2 x .5 x 2500 = 625 m

Aav= average velocity and I am finding displacement. Vi is initial velocity. T is time

For the second part. I did the same thing but replace average velocity to full speed velocity which is .5m/s x 50 Seconds. = 25m/s

The third part i don't know..

 

[SammyS]

Homework Statement

A subway trains starts from station A with a uniform acceleration of .5m/s and attains full speed after 50 seconds. It travels at full speed for 120 seconds then is brought to rest at station by by the application of brakes which provides a acceleration of -2.50.

Find the distance between A and B

Homework Equations

The Attempt at a Solution

I divided this solution into 3 stages.

1. Acceleration
2. Constant
3. deceleration

For the first part, I used Δd=viΔt + 1/2 aav Δt^2

→ 0 x 50 + 1/2 x .5 x 2500 = 625 m

Aav= average [STRIKE]velocity[/STRIKE] acceleration and I am finding displacement. Vi is initial velocity. T is time

The first part is correct (except for the typo).

For the second part. I did the same thing but replace average velocity to full speed velocity which is .5m/s x 50 Seconds. = 25m/s

For the second part, what did you use for average acceleration ?

25m/s is the initial velocity for this part.

The third part i don't know..

The initial velocity is the same as for part 2.

The acceleration is given. how long does it take to go from 25m/s to 0 m/s at this acceleration?

 

[kencamarador]

The first part is correct (except for the typo).

For the second part, what did you use for average acceleration ?

25m/s is the initial velocity for this part.

The initial velocity is the same as for part 2.

The acceleration is given. how long does it take to go from 25m/s to 0 m/s at this acceleration?

Wait sorry. 25 is the average acceleration. I meant acceleration not velocity. So I used the same equation as step 1 but replaced average acceleration with 25

 

[CAF123]

You could also consider sketching a v-t diagram and extracting the distance from there.

 

[kencamarador]

You could also consider sketching a v-t diagram and extracting the distance from there.

I tired, but for the 3rd stage, I am only given average acceleration and and initial velocity.. How can I find the distance for the last part

 

[CAF123]

I tired, but for the 3rd stage, I am only given average acceleration and and initial velocity.. How can I find the distance for the last part

Just use a kinematic equation. The initial velocity for this part of the journey is the 'full speed' attained in part 2) of the journey. You know at the end of this part, the velocity is zero and you also know the (negative) acceleration, so what eqn can you use?

 

[kencamarador]

Just use a kinematic equation. The initial velocity for this part of the journey is the 'full speed' attained in part 2) of the journey. You know at the end of this part, the velocity is zero and you also know the (negative) acceleration, so what eqn can you use?

Ok, what would be the initial velocity for the second part?

 

[CAF123]

The full speed is attained at the end of part 1). You know the distance covered in part 1) (you calculated this) and you know the acceleration.

 

[kencamarador]

I GOT IT! 3750 meters (e)

Thank you so much! Thank you!

 

[kencamarador]

The full speed is attained at the end of part 1). You know the distance covered in part 1) (you calculated this) and you know the acceleration.

I GOT IT! 3750 meters (e)

Thank you so much! Thank you!

 

[SammyS]

Wait sorry. 25 is the average acceleration. I meant acceleration not velocity. So I used the same equation as step 1 but replaced average acceleration with 25

For the first part of the trip:

The acceleration is uniform (constant) and it's 0.5 m/s², so that is also the average acceleration for this part of the trip.

You have the right numbers in the kinematic equation,

Δd=v_iΔt + (1/2) a_av (Δt)²

when you wrote

0 x 50 + 1/2 x .5 x 2500 = 625 m

even though you referred to some of the numbers erroneously.

Then you found the final velocity for the first part, although you called it the second part.

v_f = v_i (Δt) → 0m/s + 0.5m/s x 50 Seconds. = 25m/s

Note: This is also the uniform velocity for the second part.​

The second part:

The velocity is constant. The acceleration is zero.

How far does the train go in 120 seconds at 25 m/s ?​

Part 3:

What is the initial velocity for this part?

What is the final velocity?

Ehat is the acceleration?​