### Video Transcript

A particle moves along the π₯-axis. Its velocity at time π‘ is given by π£ of π‘ is equal to π‘ minus π times π‘ minus π, where π and π are constants such that π does not equal π. For which of the following values of π‘ is the particle at minimum velocity? a) π‘ is equal to ππ, b) π‘ is equal to π plus π over two, c) π‘ is equal to π plus π, or d) π‘ is equal to π and π‘ is equal to π.

Weβre given the velocity of a particle is a function of π‘. π£ of π‘ is equal to π‘ minus π times π‘ minus π, where π and π are constants such that π does not equal π. And weβre asked to find which value of π‘ is the particle at minimum velocity. Remember that we can use differentiation to find where a function has a stationary point and a minimum is a stationary point. In order to find a stationary point, we differentiate and find where the derivative is equal to zero. And that gives us the location of our stationary point. So we need to solve dπ£ by dπ‘ is equal to zero.

But first we have to find dπ£ by dπ‘ . If we multiply out the brackets for our function, we find the π£ of π‘ is equal to π‘ squared minus π plus π times π‘ plus ππ. This makes it easier to differentiate. We now have three terms to differentiate separately. So we have dπ£ by dπ‘ is equal to d by dπ‘ of π‘ squared plus d by dπ‘ of negative π plus π times π‘ plus d by dπ‘ of ππ. Since ππ is a constant, our final term d by dπ‘ of ππ is equal to zero. Since our second term is a constant times π‘, the derivative is the constant. And for our first term, we can use the power rule. This says that d by dπ₯ of ππ₯ to π is equal to πππ₯ to the π minus one where π is a constant and π is a real number.

So the derivative of our first term is two π‘. And so we have dπ£ by dπ‘ is equal to two π‘ minus π plus π. And to find our stationary points, we need to solve dπ£ by dπ‘ is equal to zero. In our case, if dπ£ by dπ‘ is equal to zero, then we have zero is equal to two π‘ minus π plus π. Rearranging this gives us two π‘ is equal to π plus π, so that π‘ is equal to π plus π over two. This corresponds to answer b in our possibilities. But letβs just check that we really do have a minimum.

We can do this by differentiating one more time. We can check that our value is a minimum by us taking d squared π£ by dπ‘ squared. Now, dπ£ by dπ‘ is two π‘ minus π plus π where π and π are constants. We find the second derivative d squared π£ by dπ‘ squared is equal to two. This value is greater than zero. Hence, we do indeed have a minimum. And since our original function π£ of π‘ is a quadratic and therefore has only one stationary point, we can assume that this is a global minimum. Our particle then has minimum velocity at π‘ is equal to π plus π over two.