Video Transcript
A particle moves along the ๐ฅ-axis. When its displacement from the origin is ๐ metres, its velocity is given by ๐ฃ equals four over three plus ๐ metres per second. Find the particleโs acceleration when ๐ is equal to three metres.
Here, we have a function for the velocity in terms of ๐ . Weโre being asked to find the acceleration. Now, acceleration is defined as the change in velocity with respect to time or the derivative of ๐ฃ with respect to ๐ก. We canโt easily differentiate ๐ฃ with respect to ๐ก though without performing an extra step. So weโre going to use implicit differentiation. We differentiate both sides of our equation with respect to ๐ก. And on the left, we get d๐ฃ by d๐ก. On the right, we get d by d๐ก of four over three plus ๐ .
To make this easier, weโre going to change this to four times three plus ๐ to the power of negative one. And then the derivative of this expression with respect to ๐ก is equal to the derivative of it with respect to ๐ times d๐ by d๐ก. We then use the general power rule on the extension of the chain rule. And we see that derivative of four times three plus ๐ to the power of negative one with respect to ๐ is negative four times three plus ๐ to the power of negative two. And so we see that d๐ฃ by d๐ก is equal to negative four times three plus ๐ to the power of negative two times d๐ by d๐ก. We can write three plus ๐ to the power of negative two as one over three plus ๐ squared. But we also know that d๐ by d๐ก is ๐ฃ. So we can see we have an expression for the acceleration in terms of ๐ฃ and ๐ .
We need to evaluate this when ๐ is equal to three. So letโs begin by working out ๐ฃ when ๐ is equal to three. When ๐ is equal to three, ๐ฃ is equal to four over three plus three which simplifies to two-thirds. And when ๐ is equal to three, acceleration is therefore is negative four over three plus three all squared times two-thirds which is equal to negative two over 27 metres per square second.
