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

[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 don't know if chain rule would work..and i don't know how to do it..

Thanks in advance!

 

[Feldoh]

Where are you getting a "y"?

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

 

[kring_c14]

Where are you getting a "y"?
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sorry..i typed the wrong letter...it should be "a"..

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

 

[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.

 

[Nesha]

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