 Well, firstly, we know that the acceleration a is equal to the change in velocity v divided by the change in time t. So let's try to work out what each one of these is separately. Let's start by working out the change in velocity delta v. Now, you may be able to see that the change in velocity is dependent on where along the circle the object is. For example, if we waste for the object to undergo a complete revolution and looked at it in the exact same spot, then the change in velocity will be zero because it will be the exact same velocity vector. So what we really want to do is we want to find the instantaneous change in velocity to find the instantaneous acceleration. So we're going to take a very, very tiny slice of the object's motion, infinitesimally small, to find the change in velocity at this point over the time it's taken to undergo this change. But to make our lives easier, we'll draw it far away and we'll just understand this change is actually meant to be very small. Now, the velocity at the first point along the object's path will call v1, and the velocity at the second point along the object's path will call v2. So we're after delta v, the change in velocity. This is the final velocity v2 minus the initial velocity v1. So let's draw it out. We now have a triangle of three sides, v1, v2, and delta v. To get an expression for delta v, we'll need to work out some information about the angles and the sides of the triangle v1, v2, and delta v. We'll start by calling this angle theta. Now remember that theta is really small as we work this out. Now, we're going to use some geometry to get the angle between v2 and minus v1. If we draw dashed lines out from our original v1 and v2 positions, we know that each of these dashed lines is perpendicular to the radial lines of the circle. We can find the angle in between our dashed lines by remembering that the sum of all angles in a quadrangle is 360 degrees. We know that two of these angles are 90 degrees, and the other one is theta, so the upper angle must be 180 degrees minus theta. Now we also know that minus v1 in our triangle is parallel to the dashed line extension of v1, original vector. Since angles on our line must sum to 180 degrees, we know that this angle here must be theta. And since we have two parallel lines, the angle between v2 and minus v1 will also be theta. This is known as the Alternate Angles Rule. So now we're going to zoom in on this triangle to work out an expression for theta. What else do we know about this triangle? Remember, we're considering uniform circular motion, so the speed is constant. This means that the length of v2 is equal to the length of v1, as these are both the magnitude of our velocity vector, and these are both equal to the speed v. As both of these sides are equal, we also know that our triangle is an isosceles triangle. Isosceles triangles can really help us out in problems, because if we cut it in half, we know that we'll get two right-angled triangles. This means we can use Sokotoa Ultragonometry Rule. So if we consider our diagram, v is the hypotenuse, half of delta v is the opposite side, and the angle is half theta. So we can write sine of theta on 2 is equal to delta v on 2 divided by v. We can actually simplify this expression down even further by using the small angle approximation. Remember, the theta is really, really tiny and infinitesimally small. The small angle approximation allows us to approximate sine of an angle as the angle itself. So sine of theta on 2 will be approximately equal to theta on 2. And for the limit case, which is the case we're considering, as we approach an infinitesimally small angle, sine theta on 2 will actually become theta on 2 exactly. So plugging this back in, we find theta on 2 is equal to delta v on 2 divided by v. Rearranging, we get the change in velocity delta v is equal to v times theta. Therefore, we have the first ingredient for our acceleration. The change in velocity is equal to v theta.