# Question Video: Finding the PowerβTime Expression of a Body given Its PositionβTime Expression Mathematics

A body of mass 17 kg moves under the action of a force π. Its position vector at time π‘ is given by the relation π«(π‘) = (7π‘Β³)π’ + (4π‘Β²)π£. Given that π is measured in newtons, π« in meters, and π‘ in seconds, write an expression for the power of force π at time π‘.

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### Video Transcript

A body of mass 17 kilograms moves under the action of a force π. Its position vector at time π‘ is given by the relation π« of π‘ equals 7π‘ cubed π’ plus 4π‘ squared π£. Given that π is measured in newtons, π« in meters, and π‘ in seconds, write an expression for the power of force π at time π‘. Let π― be the instantaneous velocity of the object.

We know that the rate at which the force π performs work on the mass, otherwise known as power, is given by π is equal to π dot π―. So, ideally, to be able to find the power of our force π, we need to identify what the force is and what the velocity is.

Weβre given that the position vector at time π‘ is 7π‘ cubed π’ plus 4π‘ squared π£. And we also know that the velocity is the rate of change of the displacement of an object with respect to time. So, π― is equal to dπ« by dπ‘. And this works for vector form as well. So, we can find a vector that describes the velocity of the body by differentiating each term with respect to π‘.

Letβs begin by differentiating the component for π’. For a power term like this, we simply multiply the entire term by the exponent and then reduce the exponent by one. So, we get three times 7π‘ squared, or 21π‘ squared. Similarly, differentiating 4π‘ squared, and we get two times 4π‘ to the power of one, or simply eight π‘. And so, now, we have a vector that describes the velocity of the body at π‘ seconds.

But what about the force? Well, one of the formulae we can use is force is equal to mass times acceleration. And of course, this works absolutely fine for vector quantities too. If we then recall that acceleration is the rate of change of velocity with respect to time β in other words, π is equal to the derivative of π― with respect to π‘ β we see we can find a vector that describes the acceleration of the body by differentiating the vector for velocity with respect to time.

Once again, we can do this separately for the horizontal or vertical components. The derivative of 21π‘ squared with respect to π‘ is 42π‘. And the derivative of eight π‘ is eight. Weβre told that the body has a mass of 17 kilograms, so force is equal to 17 times the vector for acceleration. Thatβs 42π‘ π’ plus eight π£. We can simply distribute 17 across our vector. 17 times 42π‘ is 714π‘. And 17 times eight is 136.

Power is the dot product of the vector for force and the vector for velocity. So, we multiply the π’-components. Thatβs 21π‘ squared by 714π‘. And we add the product of the π£-components. Thatβs eight π‘ times 136. That gives us 14994π‘ cubed plus 1088π‘. But what are the units for our power? Well, the force is measured in newtons, the displacement in meters, and the time in seconds. We know that one newton meter per second of power is equal to one watt of power. And so, our units are watts. The power of force π at time π‘ is 14994π‘ cubed plus 1088π‘ watts.