An object moves in a straight line. At time 𝑡 seconds, its velocity is given by 𝑡 cubed plus two meters per second. Its mass varies with time as three 𝑡 plus one kilograms. Find the magnitude of the force acting at time 𝑡 equals three seconds.
Remember, when looking at calculating the magnitude of a force over an object whose mass varies with time, we use the formula 𝐹 equals 𝑚 times d𝑣 by d𝑡 plus 𝑣 times d𝑚 by d𝑡, where 𝑚 is the mass of the object and 𝑣 is its velocity. In the question, we’re told that the mass of the object here varies with time as three 𝑡 plus one kilogram. And its velocity is 𝑡 cubed plus two meters per second.
Inspecting the formula that we have for the force of the object, we see that we need to differentiate each of these with respect to time. And of course we can do that term by term. So the derivative of 𝑚 with respect to 𝑡 is three, and the derivative of 𝑣 with respect to time is three 𝑡 squared. The force is then 𝑚 times d𝑣 by d𝑡, that’s three 𝑡 plus one times three 𝑡 squared, plus 𝑣 times d𝑚 by d𝑡, which is 𝑡 cubed plus two times three.
Let’s distribute the parentheses in each pair of terms. And that gives us nine 𝑡 cubed plus three 𝑡 squared plus three 𝑡 cubed plus six, which simplifies to 12𝑡 cubed plus three 𝑡 squared plus six. The question asks us to find the magnitude of the force that acts at 𝑡 equals three seconds. So we need to substitute 𝑡 equals three into our equation for 𝐹. That’s 𝐹 equals 12 times three cubed plus three times three squared plus six. That’s 324 plus 27 plus six, which is 357, and the unit here is newtons.
Since the object is moving in a straight line, this is indeed the magnitude of the force. It’s 357 newtons.