00:01
So in this question, we're first of all given the kinematic equation v equals v0 plus a t.
00:08
And we're asked how to determine this v0, the initial velocity and the acceleration.
00:25
And let's remember that if we have a straight line graph, v and t, then this is going to have a y intercept here.
00:38
Here and its gradient is, well, it's the slope of the graph.
00:45
And remember that y equals mx plus c is the equation of a line with gradient m and in y intercept c.
00:52
So here we can see clearly that when y is v and x is t, we're looking at a is the gradient and c is the initial velocity.
01:10
So this is our initial velocity, and the gradient here is a.
01:19
So the answer is b, slope is acceleration, and y intercept is the initial velocity.
01:42
Okay, so now we've got a kinematic equation, x equals v -0t plus a half a t squared.
01:53
And how is this graph going to look? well, we can see that x depends on both t and t squared.
02:04
And what that means is that we're looking at something parabolic.
02:15
So the answer is going to be c, that the graph looks parabolic...