Video Transcript
A particle is moving in a straight
line such that its velocity at time 𝑡 seconds is given by 𝑣 equals 15𝑡 squared
minus eight 𝑡 meters per second when 𝑡 is greater than or equal to zero. Given that its initial position
from a fixed point is 20 meters, find an expression for its displacement at time 𝑡
seconds.
Remember, velocity is the rate of
change of displacement of an object. This means we can differentiate a
function for displacement to find a function for velocity. Conversely, we can say that the
integral of the function in velocity will provide us with a function for
displacement. So to answer this question, we’re
going to integrate our function with respect to 𝑡 and then use the information
about the initial position to find a full expression for the displacement. 𝑠 is equal to the integral of 15𝑡
squared minus eight 𝑡 with respect to 𝑡.
Remember, the integral of the sum
or difference of two or more functions is equal to the sum or difference of the
integral of each function. So we can actually integrate 15𝑡
squared and negative eight 𝑡 individually. We know also that to integrate a
term of this form, we simply add one to the power and then divide by this new
number. And we obtain the interval of 15𝑡
squared to be 15𝑡 cubed divided by three. Now, in fact, we obtain a constant
of integration too. But we’ll deal with that in a
moment.
The integral of negative eight 𝑡
is negative eight 𝑡 squared divided by two. Then, we combine the two constants
of integration obtained by integrating 15𝑡 squared and negative eight 𝑡. And we see that 𝑠 is equal to 15𝑡
cubed over three plus negative eight 𝑡 squared over two plus 𝑐. We can simplify this somewhat. And what we’ve actually obtained is
the general equation for the displacement of our object. It’s five 𝑡 cubed minus four 𝑡
squared plus 𝑐. Now we can actually work out the
particular equation. That involves finding the value of
𝑐. And we can do that because we’re
told the initial position from a fixed point, in other words, its initial
displacement. We’re told its initial position
from the fixed point is 20 meters. So when 𝑡 is equal to zero, 𝑠 is
equal to 20.
So let’s substitute these values
into our result. We have 20 equals five times zero
cubed minus four times zero cubed plus 𝑐. Well, that just simplifies to 20
equals 𝑐. So the answer here is 𝑠 equals
five 𝑡 cubed minus four 𝑡 squared plus 20 meters. And to remember of course, we can
reverse this process and differentiate this expression with respect to 𝑡 to check
our solution. When we do, we obtain d𝑠 by
d𝑡. And that’s of course read to be
three times five 𝑡 squared minus two times four 𝑡, which simplifies to 15𝑡
squared minus eight 𝑡 as required.
