Video Transcript
A particle moves on a straight line so that its velocity at time 𝑡 is given by 𝑣 equals six root 𝑠, where 𝑠 is its distance from the origin. If 𝑠 equals zero when 𝑡 equals two-thirds, find 𝑠 when 𝑡 is equal to two.
This is quite an unusual rectilinear motion question because we’ve been given a function for 𝑣 in terms of 𝑠 where we’re used to seeing functions for 𝑣 in terms of 𝑡. So we’re going to begin by recalling the definition of velocity. It’s change in displacement with respect to time. So if we let 𝑠 be a function for displacement in terms of time, then velocity is equal to the derivative of displacement with respect to time, d𝑠 by d𝑡. We can, therefore, replace 𝑣 with d𝑠 by d𝑡 in our equation. And we see that d𝑠 by d𝑡 is equal to six times the square root of 𝑠.
This is a differential equation. By performing a process called separation of variables, we can collect together the 𝑠 terms and the 𝑡 terms. And we obtain one over the square root of 𝑠, d𝑠, to be equal to six d𝑡. Now, this is great because we can integrate both sides of our equation. We’re going to rewrite one over the square root of 𝑠 as 𝑠 to the power of negative one-half. And we see that the integral of 𝑠 to the power of negative one-half with respect to 𝑠, is equal to the integral of six with respect to 𝑡. So let’s integrate both sides of this equation.
To integrate 𝑠 to the power of negative one-half, we add one to the power that gives us one-half. And we divide by the new value. So this is 𝑠 to the power of one-half divided by one half which is, of course, equal to two times 𝑠 to the power of one-half. And we do actually get a constant of integration on the left-hand side. But we’re not going to write that just yet for reasons that will become clear. The integral of six is six 𝑡. And then, of course, we add plus 𝑐. This is a combination of both constants of integration. And so we have a general solution to the equation d𝑠 by d𝑡 equals six times the square root of 𝑠. It’s two times 𝑠 to the power of one-half equals six 𝑡 plus 𝑐.
We, however, have an initial value. We’re told that when 𝑠 is equal to zero, 𝑡 is equal to two-thirds. So let’s substitute these values into our equation. When we do, we obtain zero to be equal to six times two-thirds plus 𝑐. Well, six times two-thirds is four. So we subtract four from both sides and we see that 𝑐 is equal to negative four. And our equation is two times 𝑠 to the power of one-half equals six 𝑡 minus four. This isn’t very nice to look at. So we will rearrange to find an equation for 𝑠 in terms of 𝑡.
We begin by dividing both sides by two. And then we square both sides of our equation. And we see that the particular solution is 𝑠 equals three 𝑡 minus two squared. We’re looking to find the value of 𝑠 when 𝑡 is equal to two. So all that’s left to do is substitute 𝑡 equals two into our equation. That gives us six minus two all squared, which is of course four squared which we know to be 16. When 𝑡 is equal to two 𝑠 is equal to 16.
