Video Transcript
Two bodies, initially at rest at
the same point, start to move in the same direction along the same straight
line. At time 𝑡 seconds, where 𝑡 is
greater than or equal to zero, their velocities are given by 𝑣 sub one equals 13𝑡
minus 27 centimeters per second and 𝑣 sub two equals three 𝑡 squared minus five 𝑡
centimeters per second. Determine the time taken for the
two bodies to be 27 centimeters apart.
We’re given the velocity of the two
bodies as some function in terms of time. And we’re asked to use these
formulae to find information about the displacement of our bodies, specifically when
they’re 27 centimeters apart. Now, in fact, we know that we can
find the displacement of a body between two times, 𝑡 sub one and 𝑡 sub two, by
integrating the expression for velocity with respect to time between those two
times. Now, since the bodies start to move
at the same time, we can say that the initial time is equal to zero. And we’re looking to determine the
time taken. Let’s call that 𝑡.
For the bodies to be 27 centimeters
apart, we can say that the magnitudes of the differences between their two
displacements must be equal to 27. So either 𝑠 sub two minus 𝑠 sub
one equals 27 or 𝑠 sub one minus 𝑠 sub two equals 27. What we’ll do is determine
expressions for their displacement at time 𝑡 and then work out which of those is
likely to be larger. Specifically, if we define 𝑠 sub
one to be the displacement of the first body, which has velocity 𝑣 sub one, then
the displacement is going to be the definite integral between zero and 𝑡 of 13𝑡
minus 27 with respect to time. Using the power rule for
integration, and we get 13 over two 𝑡 squared minus 27𝑡 between zero and 𝑡. That gives us 13 over two 𝑡
squared minus 27𝑡 minus zero, or simply 13 over two 𝑡 squared minus 27𝑡. Then, we’ll integrate the
expression three 𝑡 squared minus five 𝑡 with respect to 𝑡 between the limits of
zero and 𝑡 to find an expression for the displacement of 𝑠 sub two. This time, we get 𝑡 cubed minus
five over two 𝑡 squared between the limits zero and 𝑡, which is equal to 𝑡 cubed
minus five over two 𝑡 squared.
Then, if we were to sketch the
graph of 𝑦 equals 13 over two 𝑡 squared minus 27𝑡 and 𝑦 equals 𝑡 cubed minus
five over two 𝑡 squared, we see that the graph of 𝑦 equals 𝑠 sub two will sit
above the graph of 𝑦 equals 𝑠 sub one for values of 𝑡 greater than or equal to
zero. This means in our interval 𝑡 is
greater than or equal to zero. The displacement of our second
particle is going to be greater than the displacement of our first.
So, returning to our earlier
equation, where we said the magnitude of the difference between 𝑠 sub two and 𝑠
sub one is equal to 27, we can simply replace the magnitude with 𝑠 sub two minus 𝑠
sub one. So we get 𝑡 cubed minus five 𝑡
squared minus 13 over two 𝑡 squared minus 27𝑡 equals 27. Simplifying and subtracting 27 from
both sides, and this becomes 𝑡 cubed minus nine 𝑡 squared plus 27𝑡 minus 27
equals zero. So how do we solve this for 𝑡?
Well, we can use the factor
theorem. For instance, let’s imagine we
substitute 𝑡 equals three into our function. If we get zero, then 𝑡 minus three
must be a factor of our polynomial. Well, when we substitute 𝑡 equals
three in, we do indeed get zero. Now, it’s worth noting that the
reason we chose three is because we had several factors of three here. This means 𝑡 equals three is a
solution to this equation. In fact, this is the only solution
to the equation. And we could test this if we chose
by using polynomial long division and then attempting to factor the remaining
quadratic. But we have found the time taken
for the two bodies to be 27 centimeters apart. It was three seconds.
