# Homework Help: How to graph velocity where there is a change in acceleration

1. Sep 16, 2014

### Kerrie yu

The speed is a constant of 20m/s for 100m (moving on a straight line) after 100m, the car stops at a constant 6m/s/s.
*So in this case, speed is equal to velocity?
*How do you graph velocity vs time graph and acceleration vs time graph?

2. Sep 16, 2014

### Staff: Mentor

The problem statement tells you specifically how the acceleration is varying with time. What is your quantitative understanding of this?

3. Sep 16, 2014

### Kerrie yu

But first I need to know if constant speed is equal to velocity. So if there is a constant speed of 20m/s , is it mean that the velocity is constant 20m/s ?

4. Sep 16, 2014

### Staff: Mentor

The problem implies that this is the case.

Chet

5. Sep 17, 2014

### Kerrie yu

So for the displacement vs time graph, the line would be going up from 0 to 100 ( take 5 sec) and then from 6 sec, it starts going down by 6 because of the acceleration. 5s=100m, 6s=94m and so on?

6. Sep 17, 2014

### Staff: Mentor

From 0 to 5 sec., this is correct. After five seconds, it is incorrect.

Chet

7. Sep 17, 2014

### Kerrie yu

The second one? Df=.......

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8. Sep 17, 2014

### Staff: Mentor

Yes, but you would have to restart the application of the equation after the first 5 seconds, so d in the equation would be the distance relative to the location after 5 seconds, and t in the equation would be 0 at the running time of 5 seconds.

Hope this makes sense. Now, what do you get?

Chet

9. Sep 17, 2014

### Kerrie yu

So we r starting new after 5sec? Df=100 ,vi=20, tf=0 ,a= -6 substitute these numbers?

10. Sep 17, 2014

### Kerrie yu

Wait no,,, di =100 , df is unknown

11. Sep 17, 2014

### Kerrie yu

..wait ..so since im starting another one,, everything before 5sec doesnt matter,, so my vi is not 20m/s anymore?

12. Sep 17, 2014

### Staff: Mentor

You seem to be pretty confused. But, you are beginning to get the right idea.

I might be better to think of this as two separate problems:

Problem 1:
Applies between time 0 and 5 seconds
Initial distance = 0
Velocity = 20 meters/sec
Acceleration = 0

You have already solved problem 1 correctly

Problem 2
Applies after 5 seconds
Initial distance = 100
Initial velocity = 20
Acceleration = -6

You need to modify your equations a little to handle this part:

v = 20 - 6t'
d = 100 + 20t'-6(t')2/2
where t'=(t-5)

So,

v = 20 - 6(t-5)
d = 100 + 20(t-5) - 6(t-5)2/2

Do these equations make sense to you?

Chet