Video: Finding the Expression of the Force Acting on a Body of Variable Mass given the Mass and Displacement Expressions as Functions of Time

A body moves in a straight line. At time 𝑡 seconds, its displacement from a fixed point is given by 𝑠 = (6𝑡² + 9𝑡) m. Its mass varies with time such that 𝑚 = (8𝑡 + 9) kg. Write an expression for the force acting on the body at time 𝑡.

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Video Transcript

A body moves in a straight line. At time 𝑡 seconds, its displacement from a fixed point is given by 𝑠 equals six 𝑡 squared plus nine 𝑡 meters. Its mass varies with time such that 𝑚 equals eight 𝑡 plus nine kilograms. Write an expression for the force acting on the body at time 𝑡.

We’ve been given an expression for the displacement of the body from a fixed point and its mass. Now, usually, we might be thinking that, to find an expression for the force, we use force equals mass times acceleration. The problem is the mass of the body varies with time. And so, we can’t quite use that formula. And so, since mass is time dependent, it’s a function of 𝑡, we use a different formula. So instead, we rearrange Newton’s second law and add a term to account for momentum entering or leaving the system. So we find that 𝐹 is equal to 𝑣 times d𝑚 by d𝑡, that’s the derivative of mass with respect to time, plus 𝑚 times d𝑣 by d𝑡, the derivative of 𝑣 with respect to time.

So how do we find expressions for 𝑣 and d𝑣 by d𝑡? Well, we’re told that displacement 𝑠 is six 𝑡 squared plus nine 𝑡. And we can obtain two pieces of information from this. We know that the velocity is the rate of change of displacement with respect to time. In other words, it’s the derivative of 𝑠 with respect to 𝑡. We can then differentiate our function for velocity with respect to time. And actually, what that does is it gives us a function for acceleration. So let’s differentiate six 𝑡 squared plus nine 𝑡 with respect to time, remembering, of course, that we can do this term by term.

When we differentiate six 𝑡 squared, we multiply the entire term by the exponent and reduce that exponent by one. So we get two times six 𝑡 which is 12𝑡. Then when we differentiate nine 𝑡, we get nine. We can repeat this process to find d𝑣 by d𝑡 which we, of course, know is equal to acceleration. This time, the derivative of 12𝑡 with respect to 𝑡 is simply 12. But the derivative of any constant is zero. So d𝑣 by d𝑡 and, therefore, 𝑎 is equal to 12.

Now, of course, we also know that mass is given by eight 𝑡 plus nine kilograms and we need to find d𝑚 by d𝑡. As before, we differentiate term by term. Now, the derivative of eight 𝑡 is eight, whilst the derivative of nine is zero. So we find d𝑚 by d𝑡 is equal to eight. Now, force is given by 𝑣 times d𝑚 by d𝑡. That’s 12𝑡 plus nine times eight. We then add the product of 𝑚, which is eight 𝑡 plus nine, and d𝑣 by d𝑡, which is 12. So 𝐹 is equal to 12𝑡 plus nine times eight plus eight 𝑡 plus nine times 12.

Our next job is to distribute our parentheses. 12𝑡 times eight is 96𝑡 and nine times eight is 72. Distributing our second set of parentheses, and again we get 96𝑡 plus 108. This simplifies to 192𝑡 plus 180. Now, in fact, we’re working in meters, kilograms, and seconds. So the units for force are newtons. And we see then that 𝐹 is equal to 192𝑡 plus 180 newtons.

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