Proving $s = ut + \frac{1}{2} at^2 $

Question

I have been asked to prove on a graph that $s = ut + \dfrac{1}{2} at^2 $

I know that the area of the rectangle is $ut$ but the area under the triangle is $\frac{1}{2}\times t \times (v-u)$

So total displacement is $s = ut + \frac{1}{2}\times t \times (v-u) $ so how do I get the equation above?

Thanks.

Answer 1

By area under graph we find (add your rectangular and trianguar components not multiply) that:

$$s=ut+\frac{t(v-u)}{2}$$

now $a=\frac{v-u}{t}$ so multiplying both sides by $t^2/2$ gives $\frac{at^2}{2}=\frac{t(v-u)}{2}$ and the answer follows.

Answer 2

Hint : $$a=\frac{v-u}{t}, at=v-u$$ So what is $ut+\frac{1}{2}(v-u)t?$

Answer 3

The graph does not only feature $A$ but also $\partial A$. So $$ s = ut + \dfrac{1}{2} at^2 \iff \\ v = \dot{s} \wedge v = u + a t \wedge s(0) = 0 $$ We can read and confirm $v$ from the graph but not $s(0)=0$. So $s$ is only determined up to a constant.

Answer 4

Another way is to notice that as the speed is a linear function of time, the average speed is the speed at half the time (also the average of the initial and final speeds):

$$\bar v=u+a\frac t2,$$ so that the space is

$$s=\bar vt=ut+a\frac{t^2}2.$$