# Figuring Displacement from acceleration

1. Sep 9, 2007

### deenuh20

1. The problem statement, all variables and given/known data

A car starts from rest at a stop sign. It accelerates at 2.0 m/s^2 for 6.0 seconds, coasts for 2.0 s, and then slows down at a rate of 1.5 m/s^2 for the next stop sign. How far apart are the stop signs?

2. Relevant equations

not sure.

3. The attempt at a solution

I tried to split up this problem into 3 parts: from the first stop sign to after 6s when it begins to coast, when its coasting, and when it starts decelerating at the second stop sign. I think i figured out the first few parts, but how could i figure the displacement from the 3rd part? Thanks.

Last edited: Sep 9, 2007
2. Sep 9, 2007

### mjsd

this is a constant acceleration problem

relevant equations:

$$s = ut +0.5 a t^2$$
$$s = 0.5(u+v)t$$

3. Sep 9, 2007

### bob1182006

Can you show the work you've done so far?

If you know the velocities at time t=6, t=8, t to get to stop sign. You can determine the distance. you may or may not need the final t.

edit: ok you will need the 3 times, velocities at time t=6,8. You know starting/ending velocities.

so you have 3 distances:
accelating coasting decelarating
x1 **** x2 *** x3

to get the total distance just add those 3 up.

When you show your work we can help out more.

Last edited: Sep 9, 2007
4. Sep 9, 2007

### Kurdt

Staff Emeritus
You will have to find the time it takes to come to a stop for the third part.

5. Sep 9, 2007

### deenuh20

thanks but what is "u" a variable for? plus i don't have the time it takes for it to decelerate. i only have the deceleration speed. so pretty much its just the initial velocity and acceleration that i have to work with. am i missing something?

6. Sep 9, 2007

### bob1182006

you have all you need. u stands for initial velocity in that equation.

To start off with the first part. t=0 to t=6. acceleration = 2.0m/s^2

so find the distance the car travels in that part. since the car starts from rest it's initial velocity =0.

To get to the second part you need to know the initial velocity (final velocity for the first part). So use a kinematic equation to determine that velocity and you go from there.

7. Sep 9, 2007

### Kurdt

Staff Emeritus
u is the standard (or most widely used) symbol for initial speed. v is commonly used for final speed.

8. Sep 9, 2007

### rocomath

i love these types of problems :-]

9. Sep 9, 2007

### Kurdt

Staff Emeritus
An alternative method is to draw a speed-time graph and find the area underneath the line.

10. Sep 9, 2007

### deenuh20

well for the first part of the problem, a=v/t, so i found the velocity by multiplying 2 m/s^2 by 6s and got 12 m/s. from this i got a displacement of 72m after 6 seconds.

for the second part, i used the initial velocity of 12 m/s * 2seconds and got a displacement of 24m. This now gives me 96m of displacement for the first 8 seconds.

Now, for the third part, i have acceleration as -1.5 m/s^2 and an initial velocity of 12 m/s. for the above equations, x=Vi+.5at^2, i have Vi and a, but i don't have t nor x. what am i missing. thanks again.

11. Sep 9, 2007

### Kurdt

Staff Emeritus
You will have to use another of the kinematic equations to find the time. Specifically in case you don't know: $v = u + at$

12. Sep 9, 2007

### bob1182006

you first part is wrong.

12m/s is the final velocity at time 6s.

in order to find the distance between t=0 to t=6 just do:
x=x0+ut+1/2at^2

and you know starting position/velocity.

for the 3rd part you can find t if you use v=u+at. since you know the starting velocity, deceleration, and what will the final velocity when the car stops be?

using that t you can solve for the 3rd distance and add them all up.

13. Sep 9, 2007

### deenuh20

ahh got it. thanks a lot guys!