# Can the same problem be done without calculus?

I had a problem in my physics class. I have been given this diagram of a free-falling object on a fictional planet "Newtonia": The dots represent "snapshots" of the object every second. The distances between each positions are labeled in the diagram.

From this, I had to find:

• (a) Acceleration
• (b) Instantaneous velocity at the second dot
• (c) The distance above the first dot from which the object has been dropped

My solution:

I assumed that the first dot is at $t=0$, the next dot is at $t=1$ and so on. Next, if $s(t)$ is the displacement function of time, then the following points will have to be on the function: $(0,0),(1,3.4),(2,7.8)$.

Another observation about the position function is that it will be a quadratic function like this:

$$s(t)=pt^2+qt+r$$

Using the three points from above, I can find the equation of the position function by setting up a system of three equations:

$$(0,0) \;\Longrightarrow\; (0^2)p+(0)q+r=0$$ $$(1,3.4) \;\Longrightarrow\; (1^2)p+(1)q+r=3.4$$ $$(2,7.8) \;\Longrightarrow\; (2^2)p+(2)q+r=7.8$$

From the first equation, it is evident that $r=0$. This simplifies the system into two equations:

$$p+q=3.4$$ $$4p+2q=7.8$$

Solving, we get $p=0.5$ and $q=2.9$. And now, since we have p, q, and r, we know that the position function is:

$$s(t)=0.5t^2+2.9t$$

The velocity and acceleration functions are first and second order derivatives respectively, so:

$$v(t)=t+2.9$$ $$a(t)=1$$

From here:

• (a) The acceleration is $\boxed{1m/s^2}$
• (b) The instantaneous velocity at the second dot ($t=1$) is $1+2.9=\boxed{3.9m/s}$
• (c) The velocity when the object is dropped is $0m/s$. This means that $0=t+2.9$, thus $t=-2.9s$. Plugging in to displacement function, we have $s(-2.9)=-4.205$. Therefore, the object was dropped at $\boxed{4.205m\;\,\text{above the first dot.}}$

So what's the problem?

I am in a non-calculus physics class. My teacher told us to calculate the "instantenous" velocity at the second dot in this case by simply taking the length of the first interval ($3.4m$) and dividing by $1s$. This is obviously not true, since acceleration is non-zero meaning that the result will not be instantaneous but average velocity. It is a good approximation, but not the exact answer.

Since my physics class is a typical non-calculus high school physics class, the approximate answer given by my teacher would be acceptable I guess. But I am curious if it's possible to get the exact answers without calculus in this scenario? (I tried approaching it in different ways, but got nowhere besides getting a headache.)

The change in velocity is uniform because the change in displacement is uniform over every one second time interval (1 m/s). Hence over a $0.5$ second time interval, the velocity changes by $0.5$ m/s. Also with uniform change in velocity (constant acceleration), the average velocity over a $1$ second time interval occurs at the middle of the interval, with the velocity at the end of the interval being $0.5$ m/s greater than the average velocity.

$$V(1) = V_{av} + 0.5 = \frac{s}{t} + 0.5$$

$$V(1) = \frac{3.4}{1} + 0.5 = 3.9\ \text{m/s}$$

• I didn't think of that, that's very clever observation :) Sep 13, 2018 at 12:29

Use the kinematic equations, which apply for constant acceleration. The most relevant one here is: $$x(t)=x_0+v_0t+\frac12at^2.$$

I should add though, that the question is wrong if we have constant acceleration. Yeah, wrong. The difference between the distance travelled in each second is constant (1m/s²) which isn't possible in real life, where it should instead vary quadratically. Clearly your teacher is trying to assume an extremely simplified model and it's worthless to think too much about this problem, since the scenario is not physically possible anyway.

• No, the distance traveled each second should vary linearly. The total distance traveled so far should vary quadratically, and it does in this example. Sep 13, 2018 at 7:52