How can I solve for average velocities using derivatives and integrals?

[StrandedStudent]

Stressed first year university student here, fresh out of high school. I took physics in both grade 11 and 12, and thought I had a pretty good grasp on it; that is until this week. Introduction to derivatives and integrals to get from x(t) to v(t) to a(t) and vice-versa. I have a pretty good grasp on how to obtain the derivatives and how to do the integral, I just have no clue on how to solve for average velocities (with or without constant accelerations??)

The attached file is the assignment that I am stuck on, but may have a slight understanding of how to attempt them.

My thinking:

1.a) To go from v(t) to x(t), I simply take the integral of v(t). v(t)= 5t^2 + 3t + 2 m/s
Integral of v(t)= 5/3t^3 + 3/2t^2 + 2t + constant = x(t)

Is there a way to get the constant without any other information?

x(t)= 5/3t^3 + 3/2t + 2t + constant m

b) To go from v(t) to a(t), I take the derivative.
Derivative of v(t)= 10t + 3 = a(t)

a(t)= 10t + 3 m/s^2

c) Completely lost here. Am I able to just plug in the two values of t and divide by 2? My professor said something about only being able to do this if the acceleration is constant, which it's not (is it?) due to the variable t being in the equation for acceleration.

2. My thinking is that I could take the derivative of x(t) to get v(t) and a(t) and plug in the given values for when t=0. Is this correct?

Any help would be greatly appreciated

 

[Bandersnatch]

Hi there! Welcome to PF.

tegral of v(t). v(t)= 5t^2 + 3t + 2 m/s
Integral of v(t)= 5/3t^3 + 3/2t^2 + 2t + constant = x(t)

Is there a way to get the constant without any other information?

The constants you get when integrating equations of motion have always the meaning of initial value. Initial velocity, initial displacement, etc. It's easy to see once you realize that their dimensions are the same as the dimensions of what you've got on the left hand side of the equation, or when you take t=0.

Knowing this, can you assign a concrete value to the constant based on the information provided in the assignment?

c) Completely lost here. Am I able to just plug in the two values of t and divide by 2? My professor said something about only being able to do this if the acceleration is constant, which it's not (is it?) due to the variable t being in the equation for acceleration.

This should help:
http://hyperphysics.phy-astr.gsu.edu/hbase/inttyp.html#c2

2. My thinking is that I could take the derivative of x(t) to get v(t) and a(t) and plug in the given values for when t=0. Is this correct?

But when t=0 the particle needn't be at the origin. First find the time corresponding to the position at x=0, then use derivatives.

 

[gneill]

When you want to find an average of something that varies over some time or distance the usual approach is to take the integral of the thing over the given period (or distance) and divide by the period (or distance).

$$ f_{avg} = \frac{1}{end - start} \int_{start}^{end} f(x) \; dx $$

In your case it looks like you want the average velocity over the time t = 1s to t = 2s. So the time period is 1 second and you have an equation for the velocity with respect to time...

 

[StrandedStudent]

Thank you both for the replies.

I'm still kind of unsure of what to do. I thought that I may have worked out the answers, but now after reading these posts, I'm almost certain they're wrong. My previous calculus knowledge is not very thorough (I took a 1/2 credit class in high school and only learned Limits, and my university professor simply taught our class how to do the chain rule to find the derivative and how to use the exponent rule thing to find the integral, x^n+1 / n+1... He taught us these rules in under 5 minutes and on we went with little to no examples.) That being said, I am trying my best to grasp this.

I have never seen the integral sign with the subscript and superscript ("end" & "start" in gneill's formula). Do they mean anything, or are they simply symbols that have no effect on the equation?

Anyways, my understanding of the previous posts are:

1.a) x(t)= 5/3t^3 + 3/2t^2 +2t + constant
To get the constant, am I able to just plug in t=0? I am not given any points, other than that the particle is located at the origin initially. So if I understand correctly, my point would be (x,t) or (0,0).
0= 5(0)^3 + 3/2(0)^2 + 2(0) + constant (here is where I'm thinking I am very wrong...)
0= 0 + 0 + 0 + constant
0= constant?

Or would I have to put the point of (0,0) into the formula given at the start of the question, v(t)? By stating "the particle is located at the origin initially" did they mean on the Velocity-Time graph?

b) I'm confident that a(t)= 10t + 3 m/s^2... I hope

c) Still unsure of what to do here. I haven't understood what the "dx" and "dt" symbols mean yet (a one-time-5-minute explanation by the professor really did not explain things fully).

2. Given the information provided, x(t)= 2t^2 - 5t
At x(t)=0, 0=2t^2 -5t
0= t(2t-5)
t=0, t=5/2

Then using the derivatives v(t)=4t - 5 and a(t)=4
v(0)=4(0)-5= -5 or
v(5/2)=4(5/2)-5= (20/2)-5=10-5= 5

Velocity is either -5m/s or 5m/s and acceleration is constant at 4 m/s^2.

I'm truly sorry if it seems like I haven't tried question 1 due to a lack of understanding. I've been at this for a little over an hour and a half, and I really do appreciate your knowledge.

 

[Bandersnatch]

I have never seen the integral sign with the subscript and superscript ("end" & "start" in gneill's formula). Do they mean anything, or are they simply symbols that have no effect on the equation?

That is the definite integral, as opposed to indefinite integral without those limits. Its meaning is the area under the curve between the two points indicated. Calculations are pretty straightforward. You just do your regular integration to arrive at an equation, then substitute the top limit to the new equation, and deduct from it the same equation with a substituted bottom limit.
E.g.:
$$ \int_{1}^{2} 5x \; dx = \frac{1}{2}5*2^2+C - (\frac{1}{2}5*1^2+C)$$
One thing to note, is that the constants will always cancel out in the definite integral.

I'm surprised your professor gave you the exercise without explaining the maths. Perhaps you've got it somewhere in your notes?

1.a) x(t)= 5/3t^3 + 3/2t^2 +2t + constant
To get the constant, am I able to just plug in t=0? I am not given any points, other than that the particle is located at the origin initially. So if I understand correctly, my point would be (x,t) or (0,0).
0= 5(0)^3 + 3/2(0)^2 + 2(0) + constant (here is where I'm thinking I am very wrong...)
0= 0 + 0 + 0 + constant
0= constant?

Yes, that's correct. The initial position is 0. That is, the particle is at the origin at t=0.

Or would I have to put the point of (0,0) into the formula given at the start of the question, v(t)? By stating "the particle is located at the origin initially" did they mean on the Velocity-Time graph?

That would be some weird wording. If they said "the velocity of the particle is 0 initially", then it'd be right.

b) I'm confident that a(t)= 10t + 3 m/s^2... I hope

Yes, that's correct.

c) Still unsure of what to do here. I haven't understood what the "dx" and "dt" symbols mean yet (a one-time-5-minute explanation by the professor really did not explain things fully).

dx simply indicates which variable you're going to integrate. It seems pretty redundant if you've got just one variable in the equation, but it's useful to get used to using it.

Try applying the rules for the definite integral as indicated above to solve the equation posted by gneill in post #3, and find the average velocity.

2. Given the information provided, x(t)= 2t^2 - 5t
At x(t)=0, 0=2t^2 -5t
0= t(2t-5)
t=0, t=5/2

Then using the derivatives v(t)=4t - 5 and a(t)=4
v(0)=4(0)-5= -5 or
v(5/2)=4(5/2)-5= (20/2)-5=10-5= 5

Velocity is either -5m/s or 5m/s and acceleration is constant at 4 m/s^2.

This looks correct.

 

[StrandedStudent]

Thank you so much for helping out! After a night's sleep and getting up bright and early, I think I've got it. It didn't help not knowing what half of the variables/signs were, but you guys have really helped out. Thank you so much