Finding Initial Velocity in a Car's Motion Using Graph Analysis

[fatherlewis]

Hi guys,

I'm having trouble finding a reasonable answer to the initial velocity of a car.

These are my givens:
t(s) 0 1.0 2.0 3.0 4.0 5.0
x(m)0 2.3 9.2 20.7 36.8 57.5

I have solved that the average velocity = 13.8m/s
That the average acceleration = 9.2m/s^2

How do I find the initial velocity of the car?

Thanks for your help

 

[alphysicist]

Hi fatherlewis,

Hi guys,

I'm having trouble finding a reasonable answer to the initial velocity of a car.

These are my givens:
t(s) 0 1.0 2.0 3.0 4.0 5.0
x(m)0 2.3 9.2 20.7 36.8 57.5

I have solved that the average velocity = 13.8m/s
That the average acceleration = 9.2m/s^2

Could you show how you got these numbers? They don't look right to me.

 

[ehild]

Hi guys,

I'm having trouble finding a reasonable answer to the initial velocity of a car.

These are my givens:
t(s) 0 1.0 2.0 3.0 4.0 5.0
x(m)0 2.3 9.2 20.7 36.8 57.5

I have solved that the average velocity = 13.8m/s
That the average acceleration = 9.2m/s^2

How do I find the initial velocity of the car?

Thanks for your help

How did you get 9.2 m/s^2 for the average acceleration?

Make a table for t, x, v (velocity), a (acceleration).

Determine the velocity in each second, from t = 2 to 5, which is the change of x in one second. Then calculate the acceleration from t = 3 to 5, which is the change of velocity in one second.


What kind of motion is it? If the acceleration happens to be uniform, you get the initial velocity from the formula x=v₀*t + a/2*t².

ehild

 

[fatherlewis]

This is how I got the answers:

I got the average velocity by (57.5-2.3)/(5-1)= 13.8m/s

I plotted all those points, and then I drew a tangent to 7 points (5 of which were those chosen points) and then found the slope for each of those points which gave me the instantaneous velocity at each of those points.

Here are the points I used:
A: (2, 9.2)(1.2,0) = 11.5m/s
B: (3, 20.7)(1.5,0)= 13.8m/s
C: (3.3, 26)(1.6,0)= 15.3m/s
D: (3.6,30)(1.7,0)=15.8m/s
E: (4, 36.8)(1.8,0)=16.7m/s
F: (4.6, 48)(1.9,0)=17.8m/s
G: (5, 57.5)(2.1,0)= 19.8m/s

I then plotted each of the points A-G to a graph of instantaneous velocity vs time.

So for the average acceleration I did Ax=(change in instantaneous velocity)/change in time
(19.8-11.5)/(2.1-1.2)=9.2m/s^squared

 

[alphysicist]

This is how I got the answers:

I got the average velocity by (57.5-2.3)/(5-1)= 13.8m/s

This gives the average velocity from t=1 to t=5; is that what you wanted? For the average velocity for the entire motion you would want to use t=0 to t=5.

I plotted all those points, and then I drew a tangent to 7 points (5 of which were those chosen points) and then found the slope for each of those points which gave me the instantaneous velocity at each of those points.

Here are the points I used:
A: (2, 9.2)(1.2,0) = 11.5m/s
B: (3, 20.7)(1.5,0)= 13.8m/s
C: (3.3, 26)(1.6,0)= 15.3m/s
D: (3.6,30)(1.7,0)=15.8m/s
E: (4, 36.8)(1.8,0)=16.7m/s
F: (4.6, 48)(1.9,0)=17.8m/s
G: (5, 57.5)(2.1,0)= 19.8m/s

I then plotted each of the points A-G to a graph of instantaneous velocity vs time.

So for the average acceleration I did Ax=(change in instantaneous velocity)/change in time
(19.8-11.5)/(2.1-1.2)=9.2m/s^squared

This acceleration seems too large to me. However, I don't really follow what you are doing for points A though G.

But think back to the definition of average velocity:

$$a_{\rm ave}=\frac{v_2-v_1}{t_2-t_1}$$
which means the average velocity is in terms of the instantaneous velocities at times $t_1$ and $t_2$.


If you calculate the average velocity from t=0 to t=1, that would approximately equal the instantaneous velocity for what time? what about the average velocity from t=4 to t=5, it would equal the instantaneous velocity for what time?

Once you have those instantaneous velocities at two times you can calculate the average velocity, and then determine the initial velocity at t=0 by using your kinematic equations. What do you get?

 

[fatherlewis]

This gives the average velocity from t=1 to t=5; is that what you wanted? For the average velocity for the entire motion you would want to use t=0 to t=5.



This acceleration seems too large to me. However, I don't really follow what you are doing for points A though G.

But think back to the definition of average velocity:

$$a_{\rm ave}=\frac{v_2-v_1}{t_2-t_1}$$
which means the average velocity is in terms of the instantaneous velocities at times $t_1$ and $t_2$.


If you calculate the average velocity from t=0 to t=1, that would approximately equal the instantaneous velocity for what time? what about the average velocity from t=4 to t=5, it would equal the instantaneous velocity for what time?

Once you have those instantaneous velocities at two times you can calculate the average velocity, and then determine the initial velocity at t=0 by using your kinematic equations. What do you get?

Thanks for the catch 0 would make more sense for the total average velocity which then gives me:

(57.5/5)=11.5m/s as the average velocity

The question asks me to plot the givens onto a distance over time graph, and then take the tangent lines of several points and get the slope of each of those lines and each of those slopes will be equal to the instantaneous velocity at that moment. I chose the 5 given points and then picked 2 of my own. It's by no way accurate of course, very much freehand, but I did have to look at my rough graph and figure out some points.

So A-G are those points, and I got the slope for each of those points, and so for the average acceleration I (19.8-11.5)/(2.1-1.2)=9.2m/s^squared. If it's too large maybe it's because I only found the average acceleration from 1.2 seconds to 2.1 seconds? If I use it through 0 meaning: 19.8m/s is the instantaneous velocity I get at point G which is my highest given, and 5 seconds is the amount of time elapsed up until it reaches that instantaneous velocity so it becomes:

(19.8-0)/5= 3.96m/s^squared...does that make more sense as an average acceleration? But then wouldn't that mean that my initial velocity was 0m/s?

I'm getting really confused...

EDIT:
OK going back at it and plugging in the numbers in the kinematic equation to get the initial velocity I finally end up with 1.96m/s as the initial velocity.

 

[alphysicist]

Thanks for the catch 0 would make more sense for the total average velocity which then gives me:

(57.5/5)=11.5m/s as the average velocity

The question asks me to plot the givens onto a distance over time graph, and then take the tangent lines of several points and get the slope of each of those lines and each of those slopes will be equal to the instantaneous velocity at that moment. I chose the 5 given points and then picked 2 of my own. It's by no way accurate of course, very much freehand, but I did have to look at my rough graph and figure out some points.

So A-G are those points, and I got the slope for each of those points, and so for the average acceleration I (19.8-11.5)/(2.1-1.2)=9.2m/s^squared. If it's too large maybe it's because I only found the average acceleration from 1.2 seconds to 2.1 seconds? If I use it through 0 meaning: 19.8m/s is the instantaneous velocity I get at point G which is my highest given, and 5 seconds is the amount of time elapsed up until it reaches that instantaneous velocity so it becomes:

(19.8-0)/5= 3.96m/s^squared...does that make more sense as an average acceleration? But then wouldn't that mean that my initial velocity was 0m/s?

I'm getting really confused...

EDIT:
OK going back at it and plugging in the numbers in the kinematic equation to get the initial velocity I finally end up with 1.96m/s as the initial velocity.

In my previous post I had outlined the way to get an acceleration and initial velocity that would exactly give back the original data.

However, if in this problem they specifically want you to use a graph to find the velocity, then measurement uncertainties are to be expected. For example, in your point A you found that the tangent line to the t=2 point crosses the time axis at t=1.2. The reason we are getting different answer is because the tangent line really does not cross there; however, it is sometimes difficult to judge how a tangent line should go. (For example, for point A it's really crossing at (t=1.0,x=0).)

(Also when you calculated this: (19.8-11.5)/(2.1-1.2)=9.2m/s^squared you were using the wrong times; the denominator should have been (5-2) because that's the times that you calculated those instantaneous velocities for. However from your last post I think you caught that.)

 

[fatherlewis]

In my previous post I had outlined the way to get an acceleration and initial velocity that would exactly give back the original data.

However, if in this problem they specifically want you to use a graph to find the velocity, then measurement uncertainties are to be expected. For example, in your point A you found that the tangent line to the t=2 point crosses the time axis at t=1.2. The reason we are getting different answer is because the tangent line really does not cross there; however, it is sometimes difficult to judge how a tangent line should go. (For example, for point A it's really crossing at (t=1.0,x=0).)

(Also when you calculated this: (19.8-11.5)/(2.1-1.2)=9.2m/s^squared you were using the wrong times; the denominator should have been (5-2) because that's the times that you calculated those instantaneous velocities for. However from your last post I think you caught that.)

Yeah they specifically ask for me to graph the problem.

Thanks a lot for all your help alphysicist.