# Question Video: Expressing the Displacement of a Particle in Terms of Time given the Velocity Expression Mathematics • Higher Education

A particle is moving in a straight line such that its velocity at time π‘ seconds is given by π£ = (15π‘Β² β 8π‘) m/s, π‘ β₯ 0. Given that its initial position from a fixed point is 20 m, find an expression for its displacement at time π‘ seconds.

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### Video Transcript

A particle is moving in a straight line such that its velocity at time π‘ seconds is given by π£ equals 15π‘ squared minus eight π‘ meters per second when π‘ 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 π‘ 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, weβre going to integrate our function with respect to π‘ and then use the information about the initial position to find a full expression for the displacement. π  is equal to the integral of 15π‘ squared minus eight π‘ with respect to π‘.

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 15π‘ squared and negative eight π‘ 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 15π‘ squared to be 15π‘ cubed divided by three. Now, in fact, we obtain a constant of integration too. But weβll deal with that in a moment.

The integral of negative eight π‘ is negative eight π‘ squared divided by two. Then, we combine the two constants of integration obtained by integrating 15π‘ squared and negative eight π‘. And we see that π  is equal to 15π‘ cubed over three plus negative eight π‘ squared over two plus π. We can simplify this somewhat. And what weβve actually obtained is the general equation for the displacement of our object. Itβs five π‘ cubed minus four π‘ squared plus π. Now we can actually work out the particular equation. That involves finding the value of π. And we can do that because weβre told the initial position from a fixed point, in other words, its initial displacement. Weβre told its initial position from the fixed point is 20 meters. So when π‘ is equal to zero, π  is equal to 20.

So letβs substitute these values into our result. We have 20 equals five times zero cubed minus four times zero cubed plus π. Well, that just simplifies to 20 equals π. So the answer here is π  equals five π‘ cubed minus four π‘ squared plus 20 meters. And to remember of course, we can reverse this process and differentiate this expression with respect to π‘ to check our solution. When we do, we obtain dπ  by dπ‘. And thatβs of course read to be three times five π‘ squared minus two times four π‘, which simplifies to 15π‘ squared minus eight π‘ as required.