I am just starting with calculus and kinematics. I saw this particular question:
The velocity of a particle at a distance $x$ is given by $v^2 = 4\cdot (x \sin (x) + \cos (x))$. Find the acceleration.
Now I know that $a$ = $\frac{dv}{dt}$. But here there is no t variable so shoud I differentiate it with respect to x? I don't think it will be possible as $x$ is the distance.
Thank you for your help.
Observe that by chain rule, we have that $$a=\frac{dv}{dt}=\frac{dv}{dx}\cdot \frac{dx}{dt}=\frac{dv}{dx}\cdot v=v\frac{dv}{dx}$$
I think what confused here is the difference between derivative and partial derivative. For example, $$\frac{d f(x)}{d t} = \frac{\partial f}{\partial x} \frac{d x}{d t}$$ But, $$\frac{\partial f(x)}{\partial t} = 0$$ So, in your question, you just do derivative on both side and notice that $\frac{d x}{d t} = v_x$ and $\frac{d v_x}{d t} = a$. Then, you could obtain $a$ by solve the equation.
