What's the car's acceleration at a specific point?

[Northern Cardinal]

Homework Statement

A remote controlled toy car starts from rest and begins to accelerate in a straight line. The figure below represents "snapshots" of the car's position at equal 0.5 s time intervals. (Assume the positive direction is to the right. Indicate the direction with the sign of your answer.)

Using data from t = 0.5 s to t = 1.5 s, what is the car's acceleration at t = 1.0 s?

Relevant Equations

I know that change in velocity/change in time is used for average acceleration, but I'm not sure what to use for a specific point.

I did (0.9-0.1)/(1.5-0.5) = 0.8/1 --> 0.8 m/s^2
This doesn't look right to me. Is this actually correct or do I need to solve the problem a different way?

 

[Lnewqban]

Welcome, Northern Cardinal!

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/acca.html

You could verify your answer, knowing that the car should increase its velocity by $0.8~m/s$ per each second (for a constant value of acceleration).
What should it be at 1 second, what at 2 seconds, if your answer is correct?

 

[haruspex]

I did (0.9-0.1)/(1.5-0.5) = 0.8/1 --> 0.8 m/s^2

It helps to use units throughout.
What units would you get from (0.9-0.1)m/(1.5-0.5)s?

 

[Northern Cardinal]

I realized I made a mistake when typing out what I did, I meant to just put 0.8 m/s. I still don't understand what to do with this however.

 

[Lnewqban]

Perhaps if you make two graphs (distance versus time and velocity versus time) you could see how these parameters change.
The first one will give you a curve, with the values of instantaneous velocities for each point.
The second one will be a straight line, which slope is the constant acceleration of the car.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/motgraph.html#c1

For each position of the car in the OP's schematic, there is only one value of instantaneous velocity.
The velocity of the car varies constantly with time, at a constant rate, which is the value of the acceleration you need to calculate.

 

[Northern Cardinal]

I'm not entirely sure that I did this right, but I got 0.4 m/s^2. Is this correct?

 

[PeroK]

I'm not entirely sure that I did this right, but I got 0.4 m/s^2. Is this correct?

Do you know how to use a spreadsheet, like Excel?

 

[Lnewqban]

I'm not entirely sure that I did this right, but I got 0.4 m/s^2. Is this correct?

That is correct.
Could you show us what you did?

It is important that you understand, because future problems will be based on these concepts.

 

[PeroK]

That is correct.

Are you sure about that?

 

[Lnewqban]

Are you sure about that?

I was, ...until now

 

[PeroK]

I was, ...until now

Perhaps we need to check that the problem has been correctly posed. It asks for the acceleration at $t = 1s$, but the data fits a constant acceleration profile. Should the question be: what is the velocity at $t = 1s$?

@Northern Cardinal can you confirm?

 

[Northern Cardinal]

I'm pretty sure the question is asking for the car's acceleration at t = 1s. The answer box on the website has "m/s^2" following it.

 

[PeroK]

I'm pretty sure the question is asking for the car's acceleration at t = 1s. The answer box on the website has "m/s^2" following it.

Okay. $a = 0.4 m/s^2$ is not right. If I use $s = \frac 1 2 a t^2$ with $a = 0.4 m/s^2$ and $t = 2s$ I get $s = 0.8m$, which is not the car's displacement at $t = 2s$.

Why didn't you check your answer like that? What stopped you plugging in your suggested answer to see whether it was correct?

 

[Northern Cardinal]

I just started learning physics last week, this was our second real assignment. I hadn't learned how to check my work yet.
How should I do the problem instead?

 

[PeroK]

I just started learning physics last week, this was our second real assignment. I hadn't learned how to check my work yet.
How should I do the problem instead?

Technology is your friend. In addition to calculators, you can do a lot of things on a spreadsheet; and, as you get more advanced by using computer programs or sophisticated mathematical packages. That said, I find you can do a lot in Excel to get you started.

The point is: whether you use a computer, calculator or pen and paper, you can calculate these values in each case for your suggested answer. And this is what you should be doing. The more calculations you do the better you understand what happens when you use these SUVAT equations.

I have to say I find this problem slightly confusing. The data is clearly (from experience) a constant acceleration motion. So, I don't know why you would use the data from $0.5s$ to $1.5s$ specifically. Either:
$$0.1m = \frac 1 2 a (0.5s)^2 \ \ \text{or} \ \ 0.4m = \frac 1 2 a (1s)^2$$
Would give you the value of $a$, which you can then double-check for $t = 1s$ and $t = 2s$.

Note that $a$ is constant here, but the velocity is changing with time.

 

[Northern Cardinal]

Alright, thank you for your help!

 

[jbriggs444]

I just started learning physics last week, this was our second real assignment. I hadn't learned how to check my work yet.
How should I do the problem instead?

In principle, this is a curve-fitting problem. You have three data points that you are using to fit a quadratic equation.

There are three unknowns in a quadratic equation (the "a", "b" and "c" in f(x) = ax^2 + bx + c). There are three data points that you have been asked to use (0.5, 0.1), (1.0, 0.4) and (1.5, 0.9). That's three constraints. The solution can be expected to be uniquely defined.

Lagrange interpolation is one "crank and grind" approach that could be used for this if physical intuition is not helping.

But you asked for advice on checking your work. The way you check your work is to write down that quadratic equation, possibly in the SUVAT form: $s(t) = s_0 + v_0 t + \frac{1}{2} a t^2$.

[Be careful. Even though you also have a data point for t=0, there is no guarantee that your curve through the data points at t=0.5, 1.0 and 1.5 will hit it]

Plug in the time values for your data points (0.5, 1.0 and 1.5) and see if the "s" values (0.1, 0.4 and 0.9) match up.