Homework Help: Determine acceleration when velocity is v=x^3-4x^2+7x

1. Mar 2, 2008

kring_c14

The velocity of a particle moving along the x axis is given by v=$$x^{3}$$-4$$x^{2}$$+7x whwere x is in meters and v in m/s. determine the acceleration when x=5m...

the thing i know is that i just can't differentiate v to get y, since it is a function of x..

i dont know if chain rule would work..and i dont know how to do it..

2. Mar 2, 2008

Feldoh

Where are you getting a "y"?

Just find $$\frac{dv}{dt}|_{x=5m}$$

Last edited: Mar 2, 2008
3. Mar 2, 2008

kring_c14

sorry..i typed the wrong letter...it should be "a"..

may you pls explain further.. should i differentiate the equation?

4. Mar 2, 2008

Feldoh

Well acceleration is defined as a rate of change of velocity: $$\frac{d(v)}{dt}$$.

Since acceleration is just a change in velocity, you can differentiate a velocity function to get an acceleration function.

As to differentiating the function, I don't believe you'd need to use the chain rule at all, because v(t) is not in a form of f(g(x)) where f(x) and g(x) are two functions, it's cases like v(t)=f(g(x)) where the chain rule would be applied in differential calculus.

Last edited: Mar 2, 2008
5. Mar 2, 2008

Nesha

a=dv/dt=3*x^2 - 8*x+7
a(x=5m)=42 m/s^2