Question Video: Finding the Magnitude of a Force Acting on a Variable-Mass Body given Its Position Vector as a Function of Time Mathematics

Video Transcript

The mass of a body at time 𝑡 is given by 𝑚 of 𝑡 equals two 𝑡 plus 12 kilograms, whereas its position vector is 𝐫 of 𝑡 equals two 𝑡 squared plus three 𝑡 plus 15 𝐜, where 𝐜 is a constant unit vector, the vector 𝐫 is measured in meters, and 𝑡 is measured in seconds. Find the magnitude of the force acting on the body at 𝑡 equals two seconds.

In this question, we’re looking to find the magnitude of a force acting on a body with variable mass. The mass is given as a function of time, and so the mass of the body will vary given the value of 𝑡. In these cases, the rate of change of the mass and the velocity of a body will depend on the magnitude of the force and on the mass and velocity of the body. And the formula that links these is 𝐹 equals 𝑚 times d𝑣 by d𝑡 plus 𝑣 times d𝑚 by d𝑡, where 𝑚 of course is the mass of the body and 𝑣 is its velocity.

So in order to find the magnitude of the force acting on our body, we need to establish values for 𝑚, 𝑣, d𝑣 by d𝑡, and d𝑚 by d𝑡. In fact, these are going to be expressions which we can evaluate later on at 𝑡 equals two.

𝑚 is equal to two 𝑡 plus 12. So how do we evaluate d𝑚 by d𝑡? Well, we can differentiate term by term, since the derivative of the sum of two or more terms is equal to the sum of the respective derivatives of each term. Now, in fact, the derivative of two 𝑡 is simply two, and the derivative of a constant is zero. So the rate of change of mass with respect to time is constant; it has a value of two.

But how do we find the velocity? We’re given a vector that describes the position at time 𝑡. It’s two 𝑡 squared plus three 𝑡 plus 15 𝐜. Now here 𝐜 is a constant unit vector, so we can observe that the position vector acts in one direction only. This means we can consider it alternatively as a scalar in one direction. With this in mind then, we know that the velocity is the rate of change of position with respect to time. So we need to differentiate our expression for 𝐫 to find an expression for 𝑣.

Once again, we do this term by term. The derivative of two 𝑡 squared is two times two 𝑡, which is four 𝑡. Then the derivative of three 𝑡 is three and the derivative of 15 is zero. So in the direction of the constant unit vector 𝐜, the velocity is four 𝑡 plus three. With this in mind, we’re now able to calculate an expression for d𝑣 by d𝑡.

When we differentiate four 𝑡 plus three, we’re simply left with four. So we have expressions for mass, d𝑚 by d𝑡, velocity, and d𝑣 by d𝑡. So force is mass times d𝑣 by d𝑡, which is two 𝑡 plus 12 times four. Then we add the product of 𝑣 and d𝑚 by d𝑡. And that’s four 𝑡 plus three times two. We’ll now distribute the four over this first set of parentheses and the two over the second. And that gives us eight 𝑡 plus 48 plus eight 𝑡 plus six. And that’s 16𝑡 plus 54.

And now we see that we have our equation for 𝐹. The force at time 𝑡 is 16𝑡 plus 54. In fact, this is also the magnitude of the force, since we’re only considering one direction and 𝑡 can only be positive. We now need to substitute 𝑡 equals two into this equation. When 𝑡 is two, 𝐹 is 16 times two plus 54. That’s 32 plus 54, which is 86. And so the magnitude of the force acting on the body at 𝑡 equals two seconds is 86 newtons.

At this stage, it’s worth noting that we could have calculated an expression for the velocity in vector form. This would have resulted in a final force in vector form. But again, since this is only working in one direction, we would have ended up with 86 as the magnitude.