Calculating acceleration via differentiation of velocity

Question

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.

Related: What is the difference between implicit and explicit time dependence e.g. ∂ρ∂t and dρdt? — Alfred Centauri, May 20 '16 at 10:58
Possible duplicate of How to get distance when acceleration is not constant? — John Alexiou, May 20 '16 at 14:18
See also http://physics.stackexchange.com/questions/61293/nonuniform-acceleration-due-to-rubber-rope — John Alexiou, May 20 '16 at 14:19
I was asking for the variable to differentiate, NOT THE SOLUTION to the problem. — Mission Coding, May 21 '16 at 09:36

score 3 · Answer 1 · answered May 20 '16 at 08:06

3

Observe that by chain rule, we have that $$a=\frac{dv}{dt}=\frac{dv}{dx}\cdot \frac{dx}{dt}=\frac{dv}{dx}\cdot v=v\frac{dv}{dx}$$

answered May 20 '16 at 08:06

SchrodingersCat

4,923
3
23
49

1

Actually, I don't think there was need of any downvote as it just provided a hint to OP's HW query. – May 20 '16 at 09:31

score 0 · Answer 2 · answered May 20 '16 at 10:38

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.

Calculating acceleration via differentiation of velocity

2 Answers2