A particle moves along the 𝑥-axis. Its velocity at time 𝑡 is given by 𝑣 of 𝑡 is equal to 𝑡 minus 𝑎 times 𝑡 minus 𝑏, where 𝑎 and 𝑏 are constants such that 𝑎 does not equal 𝑏. For which of the following values of 𝑡 is the particle at minimum velocity? a) 𝑡 is equal to 𝑎𝑏, b) 𝑡 is equal to 𝑎 plus 𝑏 over two, c) 𝑡 is equal to 𝑎 plus 𝑏, or d) 𝑡 is equal to 𝑎 and 𝑡 is equal to 𝑏.
We’re given the velocity of a particle is a function of 𝑡. 𝑣 of 𝑡 is equal to 𝑡 minus 𝑎 times 𝑡 minus 𝑏, where 𝑎 and 𝑏 are constants such that 𝑎 does not equal 𝑏. And we’re asked to find which value of 𝑡 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 d𝑣 by d𝑡 is equal to zero.
But first we have to find d𝑣 by d𝑡 . If we multiply out the brackets for our function, we find the 𝑣 of 𝑡 is equal to 𝑡 squared minus 𝑎 plus 𝑏 times 𝑡 plus 𝑎𝑏. This makes it easier to differentiate. We now have three terms to differentiate separately. So we have d𝑣 by d𝑡 is equal to d by d𝑡 of 𝑡 squared plus d by d𝑡 of negative 𝑎 plus 𝑏 times 𝑡 plus d by d𝑡 of 𝑎𝑏. Since 𝑎𝑏 is a constant, our final term d by d𝑡 of 𝑎𝑏 is equal to zero. Since our second term is a constant times 𝑡, the derivative is the constant. And for our first term, we can use the power rule. This says that d by d𝑥 of 𝑎𝑥 to 𝑛 is equal to 𝑎𝑛𝑥 to the 𝑛 minus one where 𝑎 is a constant and 𝑛 is a real number.
So the derivative of our first term is two 𝑡. And so we have d𝑣 by d𝑡 is equal to two 𝑡 minus 𝑎 plus 𝑏. And to find our stationary points, we need to solve d𝑣 by d𝑡 is equal to zero. In our case, if d𝑣 by d𝑡 is equal to zero, then we have zero is equal to two 𝑡 minus 𝑎 plus 𝑏. Rearranging this gives us two 𝑡 is equal to 𝑎 plus 𝑏, so that 𝑡 is equal to 𝑎 plus 𝑏 over two. This corresponds to answer b in our possibilities. But let’s 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 𝑣 by d𝑡 squared. Now, d𝑣 by d𝑡 is two 𝑡 minus 𝑎 plus 𝑏 where 𝑎 and 𝑏 are constants. We find the second derivative d squared 𝑣 by d𝑡 squared is equal to two. This value is greater than zero. Hence, we do indeed have a minimum. And since our original function 𝑣 of 𝑡 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 𝑡 is equal to 𝑎 plus 𝑏 over two.