A particle moves along the 𝑥-axis. When its displacement from the origin is 𝑠 metres, its velocity is given by 𝑣 equals four over three plus 𝑠 metres per second. Find the particle’s acceleration when 𝑠 is equal to three metres.
Here, we have a function for the velocity in terms of 𝑠. We’re being asked to find the acceleration. Now, acceleration is defined as the change in velocity with respect to time or the derivative of 𝑣 with respect to 𝑡. We can’t easily differentiate 𝑣 with respect to 𝑡 though without performing an extra step. So we’re going to use implicit differentiation. We differentiate both sides of our equation with respect to 𝑡. And on the left, we get d𝑣 by d𝑡. On the right, we get d by d𝑡 of four over three plus 𝑠.
To make this easier, we’re going to change this to four times three plus 𝑠 to the power of negative one. And then the derivative of this expression with respect to 𝑡 is equal to the derivative of it with respect to 𝑠 times d𝑠 by d𝑡. We then use the general power rule on the extension of the chain rule. And we see that derivative of four times three plus 𝑠 to the power of negative one with respect to 𝑠 is negative four times three plus 𝑠 to the power of negative two. And so we see that d𝑣 by d𝑡 is equal to negative four times three plus 𝑠 to the power of negative two times d𝑠 by d𝑡. We can write three plus 𝑠 to the power of negative two as one over three plus 𝑠 squared. But we also know that d𝑠 by d𝑡 is 𝑣. So we can see we have an expression for the acceleration in terms of 𝑣 and 𝑠.
We need to evaluate this when 𝑠 is equal to three. So let’s begin by working out 𝑣 when 𝑠 is equal to three. When 𝑠 is equal to three, 𝑣 is equal to four over three plus three which simplifies to two-thirds. And when 𝑠 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.