# Question Video: Determine the Force Acting upon a Body That Moves in a Straight Line and Whose Mass Varies with Time Mathematics

A body moves in a straight line. At time π‘ seconds, its displacement from a fixed point is given by π  = (2π‘Β² + 5π‘ + 4) m. Its mass varies with time such that π = (6π‘ + 5) kg. Determine the force acting upon the body when π‘ = 3π .

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### Video Transcript

A body moves in a straight line. At time π‘ seconds, its displacement from a fixed point is given by π  equals two π‘ squared plus five π‘ plus four meters. Its mass varies with time such that π is equal to six π‘ plus five kilograms. Determine the force acting upon the body when π‘ equals three seconds.

In this question, weβve been given information about the displacement of a body moving in a straight line with respect to time. Weβve also been given information about its mass with respect to time π‘. And so we have a body with variable mass. This means we canβt use the equation πΉ equals ππ that weβre so used to.

Instead, we use Newtonβs second law, but we add a term to account for the momentum carried by mass entering the system. We say πΉ is equal to π£ times dπ by dπ‘ β thatβs the derivative of mass with respect to time β plus π by dπ£ by dπ‘, which is the derivative of π£ with respect to time, in other words its acceleration.

Now since π is equal to six π‘ plus five, weβre going to need to differentiate that with respect to time. And we can do that term by term. The derivative of six π‘ is just six, and the derivative of five is zero. So we so find that dπ by dπ‘ is equal to six.

Now weβre told that π , the displacement, is two π‘ squared plus five π‘ plus four. And so we need to recall that velocity is the change in displacement with respect to time. Thatβs dπ  by dπ‘. So we differentiate each term with respect to time. The derivative of two π‘ squared is two times two π‘, which is four π‘. And the derivative of five π‘ is five. The derivative of four, however, is zero. So we find when we differentiate π  with respect to π‘, we get four π‘ plus five. Similarly, we need dπ£ by dπ‘ for the formula. So weβre going to differentiate once again. And of course, in fact, that gives us acceleration.

Now the derivative of four π‘ is four, and the derivative of five is zero. Letβs substitute everything we have into our formula. π£ times dπ by dπ‘ is four π‘ plus five times six, and π times dπ£ by dπ‘, in other words ππ, is six π‘ plus five times four. We distribute each of our parentheses. Four π‘ times six is 24π‘, and five times six is 30. Similarly, six π‘ times four is 24π‘, and five times four is 20. And so we collect like terms, and we find that the force is 48π‘ plus 50. And since weβre working meters, seconds, and kilograms, itβs actually 48π‘ plus 50 newtons.

Now, of course, weβre looking to find the exact force at π‘ equals three. So πΉ equals 48 times three. We replace π‘ with three and then we add 50. That gives us 194. And so the force acting on the body when π‘ equals three seconds is 194 newtons.