Video Transcript
A particle is moving in a straight line such that its speed 𝑣 measured in meters per second and its position 𝑥 measured in meters satisfy the equation 𝑣 squared equals 33 minus three cos 𝑥. Find the maximum speed of the particle 𝑣 max and the acceleration of the particle 𝑎 when equals 𝑣 max.
Now, usually when we think about maxima and minima, we think about differentiating a function. We don’t need to do that here. We’re given an equation 𝑣 squared equals 33 minus three cos 𝑥. And we need to find the maximum speed of the particle 𝑣 max. And so, we’ll begin by thinking about the primary part of this function, the cos 𝑥 part. Let’s think about what we know about cos of 𝑥. The range of cos 𝑥 is values greater than or equal to negative one and less than or equal to one. So we need to think about which values of cos of 𝑥 will make 33 minus three cos 𝑥 as large as possible.
Well, if we multiply negative three by negative one, we’re going to get a positive value for negative three cos 𝑥. And so, the maximum value for 𝑣 squared and therefore 𝑣 will occur when cos 𝑥 is equal to negative one. When this happens, our expression for 𝑣 squared is 33 minus three times negative one, which is 33 plus three or 36. And so, the maximum value of 𝑣 squared is 36. To solve this equation for 𝑣, we need to perform the inverse operation to squaring. We’re going to square root both sides of the equation.
Usually, we’d consider both the positive and negative square root, but we’re just interested in speed, and speed doesn’t have a direction. It can only be positive. And so, we can say that the maximum value of 𝑣 will be equal to the square root of 36, which is simply six. We’re working in meters per second, so 𝑣 max is six meters per second.
This question also wants us to find the acceleration, 𝑎, of the particle when the speed is equal to 𝑣 max; it’s equal to six meters per second. And so, let’s consider what we know about the maximum speed of the particle. We know that the acceleration is the rate of change of velocity. And so when we think about rate of change, we think about the derivative. So, the acceleration is found by differentiating the function for 𝑣 with respect to 𝑡.
Similarly, we can find the location of any relative maxima or minima of a function by setting the derivative equal to zero. In this case, the maximum speed of the particle will occur when its first derivative is equal to zero. We said, though, that the first derivative of 𝑣 with respect to 𝑡 is 𝑎. And so, the maximum speed of the particle must occur when the acceleration itself is equal to zero. The acceleration will be in meters per second per second or meters per square second.
And so we see the maximum speed of our particle 𝑣 max is six meters per second and its acceleration is zero meters per square second.
