Question

Starting from the one-dimensional motion equation x=Xo + vt

prove that v^2 = Vo^2 + 2a(X-Xo)

If you could eplain as well why/ how each step in the problem proves the equation, this would be greatly helpful.

Thank you!

Answer 1

As I previously said, the gradient of the line is equal to acceleration a. So a=(v-u)/t. Rearranging this to make v the subject, gives us our first constant acceleration formula:

s=ut+1/2at²

We know that the area under the graph is equal to the displacement. So we know that u multiplied by t gives us the bottom rectangle of the area and v-u divided by 2, gives us the top triangle. This gives us:

s=ut+(v-u)t/2.

Now we already know that v=u+at so we can rearrange that to give v-u=at and then substitute this into our equation for displacement. From this we have s=ut+(at)t/2.

If we just multiply out the bracket that provides us with our second formula:

We've already established that the area under the graph (equal to displacement s) is equal to:

s=ut+1/2(v-u)t

If we multiply out the bracket we get:
s=ut+1/2vt-1/2ut

which is the same as:

s=1/2ut+1/2vt

Finally we just factorise this to give:

v²=u²+2as

We can rearrange v=u+at, to make t the subject:

t=(v-u)/a

Then we just substitute this value of t into our previous equation: s=1/2(u+v)t, which gives us:

s=1/2(u+v)(v-u)/a

2as=(v+u)(v-u)
v²-u²=2as