Finding average acceleration by combining constant velocity vectors

In summary, the given data is a graph depicting the distance (km) that a bus travels over 3.5 hours, broken into three linear segments. The goal is to find the average acceleration over the entire trip in km/hr^2, with the answer being negative. The first segment shows the bus traveling 25km in 1 hour, with an average velocity of 25 km/hr and an average acceleration of 50 km/hr^2. The second segment shows the bus traveling 7.25 km in 1.25 hours, with an average velocity of 6 km/hr and an average acceleration of -73.26 km/hr^2. The third segment shows the bus traveling -5 km in 1.75 hours
  • #1
kmb11132
13
0

Homework Statement



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

Homework Equations


(Unknown)

The Attempt at a Solution



(25-15.2-5.714)/3 = 1.362 km/hr^2 :frown:
 
Last edited:
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  • #2
kmb11132 said:

Homework Statement



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.

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 1.75 hours
velocity: -4 km/hr
avg. acceleration: -5.714 km/hr^2

Homework Equations


(Unknown)

The Attempt at a Solution



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

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
PeterO said:
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

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:
  • #4
kmb11132 said:
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?

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
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
kmb11132 said:
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.

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

Exactly.

k2p9J.jpg
 

What is average acceleration?

Average acceleration is the rate at which an object's velocity changes over a given period of time. It is a vector quantity, meaning it has both magnitude and direction.

How do you find average acceleration?

To find average acceleration, you divide the change in velocity by the change in time. This can be represented by the formula: average acceleration = (final velocity - initial velocity) / time interval.

What are constant velocity vectors?

Constant velocity vectors are vectors that have the same magnitude and direction throughout their entire motion. This means that the object is moving at a constant speed and in a straight line.

How do you combine constant velocity vectors to find average acceleration?

To combine constant velocity vectors, you can use vector addition. This involves adding the individual components of the vectors to find the resultant vector, which represents the overall average velocity and direction of the object's motion.

Why is finding average acceleration important?

Finding average acceleration allows us to understand how an object's velocity is changing over time. This information can be used to analyze the motion of objects and make predictions about their future behavior. It is also a fundamental concept in physics and is used in many real-world applications, such as designing vehicles and studying the movement of celestial bodies.

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