Multiple step velocity/acceleration problem

[hannah2329]

Homework Statement

a train is moving at 26.1m/s. 12.7 sec later its speed is 15.9m/s. find the trains acceleration- -.803m/s^2

what additional time would be necessary to bring the train to a stop if it continues to accelerate at the same rate calculated?
-.803=0-15.9/t which is 19.8 seconds

this is the part i can't figure out. i only have two tries left.

find the total distance necessary to bring the train to a complete stop from the beginning initial velocity.

Homework Equations

The Attempt at a Solution

i tried setting it up 26.1=d/32.5 but that wasnt right and now i don't know what else to do.

 

[Stonebridge]

You need to add the two values of distance for the two separate accelerations.
Does the fact that distance traveled = average velocity x time help?
For both accelerations you have the initial and final velocity as well as the time.
[average velocity = 0.5 x (initial + final) ]

 

[hannah2329]

I understand that I'm just not sure which velocity I'm supposed to use on each step that's where I'm stuck because everytime I tried it I got the answer wrong.

 

[Stonebridge]

a train is moving at 26.1m/s. 12.7 sec later its speed is 15.9m/s. find the trains acceleration- -.803m/s^2

1st step
velocity goes from 26.1 to 15.9m/s. What is the average?
Distance traveled = average velocity x time

what additional time would be necessary to bring the train to a stop if it continues to accelerate at the same rate calculated?
-.803=0-15.9/t which is 19.8 seconds

2nd step
Velocity goes from 15.9m/s (from part 1) to zero
What is the average velocity?
Distance is average velocity x time.

 

[rwisz]

I would first clarify that the units for velocity are typically meters per second (m/s) and for acceleration are meters per second squared (m/s^2).

To solve this problem, we can first use the formula for average acceleration, which is change in velocity divided by change in time (a=Δv/Δt), to find the acceleration of the train. We are given the initial velocity (26.1 m/s) and the final velocity (15.9 m/s), as well as the time interval (12.7 seconds). Plugging these values into the formula, we get:

a = (15.9 m/s - 26.1 m/s) / (12.7 s)
a = -10.2 m/s / 12.7 s
a = -0.803 m/s^2

So, the acceleration of the train is -0.803 m/s^2.

To find the additional time needed to bring the train to a stop, we can use the formula for final velocity (v = u + at), where u is the initial velocity, a is the acceleration, and t is the time. We know that the final velocity will be 0 m/s (since the train is coming to a stop) and we can use the acceleration we just calculated (-0.803 m/s^2). Plugging in these values, we get:

0 m/s = 15.9 m/s + (-0.803 m/s^2) * t
-15.9 m/s = -0.803 m/s^2 * t
t = -15.9 m/s / (-0.803 m/s^2)
t = 19.8 seconds

So, it would take an additional 19.8 seconds for the train to come to a complete stop if it continues to accelerate at the same rate.

To find the total distance necessary to bring the train to a complete stop, we can use the formula for displacement (s = ut + 1/2at^2), where u is the initial velocity, a is the acceleration, and t is the time. Again, we know that the final velocity will be 0 m/s and we can use the acceleration we calculated earlier (-0.803 m/s^2). Plugging in these values, we get:

s = (26.1 m/s)(19.8 s) + 1