# Establish traveled distance from velocity knowing constant acceleration

This must a simple math problem, but i'm scratching my head here.

An object is falling from its resting position with constant acceleration $$9.8m/s^2$$ (gravity) and hits the ground with velocity $$29.4m/s$$. I need to know from what distance this object fell.

I know that "flight-plan" was like this:

$$\begin{array}{c|c|c|} \text{Time (s)} & \text{Acceleration (m/s^2)} & \text{Velocity (m/s)} \\ \hline \text{1} & 9.8 & 9.8 \\ \hline \text{2} & 9.8 & 19.6 \\ \hline \text{3} & 9.8 & 29.4 \\ \hline \end{array}$$

I can establish time traveled by dividing velocity by acceleration: $$29.4 / 9.8 = 3s$$. To check the units: $$(m/s) / (m/s^2) = (m / s * s^2) / (m) = (m *s) / (m) = s$$ - seems okay. Now i need to transform time and acceleration into distance. Can i do that?

There are some pages on the Internet that suggest this formula: $$s=v_0t + 1/2at^2$$

$$s$$ - distance, $$v_0$$ - initial velocity, $$t$$ - time, $$a$$ - acceleration. In this case $$v_0$$ is zero, so it can be simplified to $$s=1/2at^2$$

But this can't be right, because if i plug values into it when $$t = 1$$, then i get $$1/2 * 9.8 * 1^2 = 1/2 * 9.8 = 4.9m$$, but it has to be $$9.8m$$ if time is $$1s$$ and acceleration is $$9.8m/s^2$$. No? If i put other values, then i also don't get result that i expect.

Is there an actual formula to get traveled distance from time and constant acceleration?

You're going wrong in saying that the body must cover $$9.8 m$$. The body only gains a velocity of $$9.8 m/s$$ over the course of $$1 s$$. Because the body gains a velocity of $$9.8 m/s$$ only just after one second it should be clear to you that the distance covered in $$1s$$ must be less than $$9.8m$$ and so is the case as you have verified

• Indeed, and since we have uniform acceleration from 0m/s to 9.8m/s, the average velocity is just the average of those two values (4.9m/s). The distance covered is the average velocity (4.9m/s) times the duration (1s), for a result of 4.9m. Apr 28, 2020 at 20:06
• @NuclearWang True that! Apr 28, 2020 at 20:10

The acceleration is the change in velocity.

During the first second, the velocity increases from zero to $$9.8$$ m/s.

The distance traveled during the first second, therefore, must be less than $$9.8$$ m.

The formula you have is correct.

• Thank you for explaining this. Indeed, at t=0 velocity was zero and at t=1 velocity was 9.8m/s^2, therefore average velocity during that time was 4.9m/s and object traveled 4.9m during first second. Apr 28, 2020 at 20:13
• @plasmatron exactly.
– John
Apr 28, 2020 at 21:03

No, it does not have to be $$9.8$$ m. Here, $$u=0, v=29.4, a=9.8$$

Using $$v^2 - u^2 = 2aS$$ we get $$(29.4)^2-0=2\cdot 9.8 \cdot S$$ or $$S=44.1 m$$