# Condition of acceleration when using $x=vt$

So my teacher told me that since $v = \delta x/ \delta t$, $\delta x = v • \delta t$ (naturally), and that is equal to the "area under the velocity-time graph", or displacement. This all makes sense to me, except that he told me that I could only use $\delta x = v • \delta t$ when the acceleration is $0$. Shouldn't it be rather only when the acceleration is constant (so velocity is linear), not necessarily $0$?

• Have you tried to work out an expression for $x$ (and $v$ ) with constant acceleration ? Start there. – StephenG Sep 3 '17 at 9:24

## 2 Answers

You have the expression for displacement as a function of time, $t$ - $$\mathbf{x}(t) = \mathbf{u}t + \dfrac{1}2 \mathbf{a}t^2$$ Where $\mathbf{u}$ is the initial velocity, $\mathbf{a}$ is the acceleration of the particle.

Set $\mathbf{a} = 0$, and you have $\mathbf{x}(t) = \mathbf{u}t$. This is the case of uniform velocity, not acceleration. You are confusing the terms, as only when velocity, $\mathbf{v} = \text{Constant/Uniform} \implies \mathbf{a} = \dfrac{d\mathbf{v}}{dt} = 0$. So this condition wouldn't correspond to uniform acceleration.

Yes your teacher is right. In that small interval of time acceleration is taken zero. Although for a large interval you can find displacement by integrating your expression when there is either constant or variable acceleration.