TEST tomorrow in physics and on this problem

[john16O]

Homework Statement

A light-rail train going from one station to the next on a straight section of track accelerates from rest at 1.4 { m/s^2} for 15 { s}. It then proceeds at constant speed for 1100 { m} before slowing down at 2.2 { m/s^2} until it stops at the station.

What is the distance between the stations?

How much time does it take the train to go between the stations?

Homework Equations

(v_x)_f= (v_x)_i + a_x$$\Delta$$t
x_f= x_i + (v_x)_i$$\Deltat$$ + 1/2a_x($$\Deltat$$)²

(v_x)_f²= (v_x)_i²+2a_x$$\Deltax$$

The Attempt at a Solution

I already know the answer but I can't figure out how to get to the right answer...
d= 1400m
t=77s

 

[jessdragon]

Split the question into three different parts. Use equations Vf = Vo + at, x= 1/2(Vf + Vo)t.

For the first part, you are given Vo, a, t. You need to first find Vf, then x.

For the second part, you are given x and constant v. You must find t. Think about the units.

For the third part, you are given a (remember, it's deccelerating so it's - ), Vo, and Vf. Find t, then x.

At the end, add up all the times from the 3 parts to get 77s. And add the displacements together to get 1400m (I got 1357.5m, but 1400m is correct for the number of sig figs).

I hoped that helped, and that it wasn't too much information on how to solve the question.

 

[baba89]

As a scientist, my response would be as follows:

Based on the information provided, we can use the equations for constant acceleration to determine the distance between the two stations and the time it takes for the train to travel between them. The initial velocity (vxi) is 0 m/s, and the final velocity (vxf) after acceleration is 21 m/s. Using the first equation, we can calculate the distance traveled during the acceleration phase:

vxf = vxi + axΔt
21 m/s = 0 m/s + 1.4 m/s^2 * 15 s
d = vxiΔt + 1/2ax(Δt)^2
d = 0 m/s * 15 s + 1/2 * 1.4 m/s^2 * (15 s)^2
d = 105 m + 157.5 m
d = 262.5 m

Next, we can use the second equation to calculate the distance traveled during the constant speed phase:

vxf^2 = vxi^2 + 2axΔx
(21 m/s)^2 = (0 m/s)^2 + 2 * 2.2 m/s^2 * Δx
Δx = (21 m/s)^2 / (4.4 m/s^2)
Δx = 441 m

Finally, we can determine the total distance between the two stations by adding the distances traveled during the acceleration and constant speed phases:

d = 262.5 m + 441 m
d = 703.5 m or 0.7035 km

To calculate the time it takes for the train to travel between the two stations, we can use the third equation:

(vx)f = (vx)i + axΔt
0 m/s = 21 m/s + (-2.2 m/s^2) * Δt
Δt = 21 m/s / 2.2 m/s^2
Δt = 9.545 s

Therefore, the total time for the train to travel between the two stations is:

t = 15 s + 9.545 s + 77 s
t = 101.545 s or 1.69 minutes

In conclusion, the distance between the two stations is 703.5 m and the time it takes for the train to travel between them