# How can you find average velocity from acceleration and time?

This is a question I came across in a lab report for my first high school physics course. And I'm a little stumped by it. So far the only way I know to get the average velocity is to add the final and initial velocities and divide by 2. Looking through my textbook and searching online I've come up empty so far. so I've come here for some answers. This is assuming constant acceleration and disregarding air resistance.

• Did u study calculus? My assumption is you didn’t. And I also wonder by other answer cause they are speaking with calculus while it will be completely meaningless to speak using calculus who doesn’t know calculus. Nov 28, 2021 at 10:30
• @BillyIstiak Calm down. One can understand the result of a calculus computation w/o actually being able to carry out that computation. And ultimately these things can only be understood by means of calculus.
– Gert
Nov 28, 2021 at 10:34

Assuming the acceleration $$a$$ is uniform, then:

$$v(t)=v_0+a\Delta t$$

where $$v_0$$ is the velocity at $$t=0$$.

If by average velocity $$\mathbb{v}$$ we mean average over time then strictly speaking:

$$\mathbb{v}=\frac{1}{\Delta t}\int_0^{\Delta t}v(t)\mathrm{d}t$$ $$=\frac{1}{\Delta t}\int_0^{\Delta t}(v_0+at)\mathrm{d}t$$ $$=\frac{1}{\Delta t}\left(v_0 \Delta t+\frac12 a(\Delta t)^2\right)$$ $$\mathbb{v}=v_0+\frac12 a \Delta t$$

If $$v_0=0$$, then $$\mathbb{v}=\frac12 a \Delta t$$

Assuming that your acceleration is constant, then you can simply use $$\bar v = \dfrac{v_1+v_2}2$$, and you can calculate both $$v_1$$ and $$v_2$$ by using the equation $$v=at+v_0$$.

Do note that if the acceleration is not constant, then the average acceleration is $$\bar v = \dfrac{\Delta x}{\Delta t}$$ where $$\Delta x$$ is the change in position.

Actually the formula which you are using will be applicable for constant acceleration also.

Let initial velocity will be $$u$$ and constant acceleration be $$a$$ then distance travelled in $$t$$ seconds will be: $$S=ut+\frac{1}{2}at^2$$

At root level average velocity means total displacement (not distance) divided by total time taken. So $$V_{avg} = \frac{X_f-X_i}{t_f-t_i}$$ From above both equations : $$V_{avg} = \frac{ut + \frac{1}{2}at^2}{t} = u + \frac{1}{2}at$$ Since, $$at = v-u$$ therefore: $$V_{avg}=\frac{u+v}{2}$$

Hope this helps