# Finding average acceleration by combining constant velocity vectors

1. Sep 14, 2011

### kmb11132

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

The given data is a graph depicting the distance (km) that a bus travels over 3.5 hours. The trip is broken up into three linear segments. I am supposed to find the average acceleration over the entire trip in km/hr^2. I am told that the answer will be negative.

Edit: The graph I have been given hasn't been made easy to read. The numbers I give below don't coincide with the ones I used when I originally calculated the velocity and acceleration. I'm recalculating those numbers now and attempting to find the solution again.

Segment 1 (from rest): 25 km in 1 hour
velocity: 25 km/hr
avg. acceleration: 25 km/hr^2

Segment 2: 7.25 km in 1.25 hours
velocity: 6 km/hr
avg. acceleration: -15.2 km/hr^2

Segment 3: -5 km in 1.25 hours
velocity: -4 km/hr
avg. acceleration: -5.714 km/hr^2

2. Relevant equations
(Unknown)

3. The attempt at a solution

(25-15.2-5.714)/3 = 1.362 km/hr^2

Last edited: Sep 14, 2011
2. Sep 14, 2011

### PeterO

Are all those values given to you, or have you calculated some of them? I don't think they line up.

Take first hour:

Bus travels 25 km in one hour

average vel = 25 km/hr [i like that]

If acceleration was constant throughout, the final vel will have been 50 km/hr

That means the acceleration will have to have been 50 km/hr2

3. Sep 14, 2011

### kmb11132

Yes, I attempted to calculate the velocities and average speed. If average acceleration is:

(v1-v0)/(t1-t0) then I thought that would be (25km-0km)/(1hr-0hr) in kilometers per hour^2 if it started from rest and traveled 25km in 1 hour.

Okay so I see that I'm doing something wrong with the units. My professor willfully leaves units out of equations and tells us not to worry about them for now so I have no idea what's wrong. As you can tell I just started Physics 101 and I'm having a little bit of a hard time. Can you tell me how the hours squared while the velocity is multiplied by two?

Last edited: Sep 14, 2011
4. Sep 14, 2011

### PeterO

If we have constant acceleration, the average velocity is merely the (initial Velocity + final velocity) / 2
So if average velocity is 25, and initial velocity is 0 then the final velocity must be 50

(0+50)/2 = 25

Hours aren't really squared - they just look like it.

Consider this:
If you sat in a car and watched the speedo as the car accelerated away from traffic lights, you might notice that it reached a speed of 60 km/h in 10 seconds.

That represents an acceleration of 6 km/h per second which would be written as km/h.s
or more likely km.h-1.s-1

If the speedometer had been calibrated in metres per second instead of km/hr, you may have seen in reach a speed of 18 m/s in 10 seconds.
That would mean 1.8 m/s per second or 1.8 m/s.s which would be written as 1.8 m/s2 or even better m.s-2

Another example. Kinetic Energy = 1/2 . m . v2 so the units are kg.m2s-2

now m2 is a standard unit for area, but there is no area involved in Kinetic energy, the m2 comes from the fact that velocity is squared; which also brings along the s-2.

Vel unit ms-1
vel2 unit m2s-2

5. Sep 15, 2011

### kmb11132

Thanks Peter, that makes a lot of sense.

So after trying to remeasure and calculate everything accurately here is what I came up with:

Segment #1: +25km in 1hr (given data)
v_i = 0km/hr
v_av = 25km/hr
v_f = 50km/hr
a_av = 50km/hr^2

Segment #2: +7.25km in 1.2hr (given data)
v_i = 50km/hr
v_av = 6.04km/hr
v_f = -37.92km/hr
a_av = -73.26km/hr^2

Segment #3: -6.25km in 1.3hr (given data)
v_i = -37.92km/hr
v_av = -4.81km/hr
v_f = 29.72km/hr
a_av = 52.03km/hr^2

Okay so I can tell I'm still not getting the numbers right. Here is what I thought:

v_i = v_f of the previous segment

v_av = (displacement of segment)/(time traveled in segment)

v_f = 2(v_av) - v_i (<- this is from manipulating v_av = (v_f - v_i)/2)

a_av = (v_f - v_i)/(time traveled in segment)

I can't tell where I'm going wrong, just that I am.

6. Sep 15, 2011

### PeterO

I really can't imagine what the graph supplied looks like, so can't really comment on your calculations.

I do know that a distance-time graph can never have a negative gradient, since there is no direction associated with distance.

If you are actually given a displacement-time graph you could have negative parts.

Does the graph consist of a slope up from the origin, a slope up a little more, then a slope down?

7. Sep 15, 2011

### kmb11132

Exactly.

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