I'm having trouble with this homework question
A mysterious rocket-propelled object of mass 49.0 kg is initially at rest in the middle of the horizontal, frictionless surface of an ice-covered lake. Then a force directed east and with magnitude $F(t) = (16.3\text{ N/s})t$ is applied.
How far does the object travel in the first 5.50s after the force is applied?
For some reason I'm not getting the correct answer. I think maybe I'm not understanding how to use the magnitude of the force they are giving me. I know how to use a constant force, but is this different because the force is a function of time?
I tried starting it like this:
$$\begin{align}F &= ma\\
16.3(t) &= (49)(a) \\
16.3(5.5) &= (49)(a) \\
89.65 &= (49)(a) \\
a &= 1.82959\ \mathrm{m/s^2}\end{align}$$
So now we know that:
$$\begin{align}t &= 5.5\text{ s} \\
a &= 1.82959 m/s^2 \\
V_o &= 0\end{align}$$
So I plug it into my equation:
$$\begin{align}\Delta X &= V_o t + 1/2 a t^2\\
\Delta X &= (0)(5.5) + (1/2)(1.82959)(5.5)^2\\
\Delta X &= 27.7\text{ m}\end{align}$$
But that's not the right answer.
Best Answer
Your mistake here is that you're using the three equations of motion ($v=u+at$ et al), even though they are only applicable for constant acceleration.
The correct way is this (i'm writing 16.3 as $k$) :$$F=ma=kt$$ $$\therefore a=kt/m$$ $$\therefore \frac{dv}{dt}=kt/m$$ $$\therefore dv=\frac{ktdt}{m}$$ $$\therefore \int\limits_0^vdv=\int\limits_0^t\frac{ktdt}{m}$$ $$\therefore v=\frac{kt^2}{2m}$$ $$\therefore \frac{dx}{dt}=\frac{kt^2}{2m}$$ $$\therefore dx=\frac{kt^2dt}{2m}$$ $$\therefore \int\limits_0^xdx=\int\limits_0^t\frac{kt^2dt}{2m}$$ $$\therefore x=\frac{kt^3}{6m}$$
I had to use a bit of calculus here. Any problem involving rate of change requires calculus. Those three equations of motion are derived from "acceleration is rate of change of velocity" and "velocity is rate of change of displacement" at constant acceleration. If $a$ is not constant, you have to use these equations: $$v=\frac{dx}{dt}$$,$$a=\frac{dv}{dt}=\frac{d^2x}{dt^2}$$. A shortcut formula for when acceleration is given in terms of displacement is $$a=v\frac{dv}{dx}$$.