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

• kring_c14
In summary, the velocity of a particle moving along the x-axis is given by v=x^{3}-4x^{2}+7x where x is in meters and v in m/s. To determine the acceleration when x=5m, you can differentiate the equation to get a=dv/dt=3*x^2 - 8*x+7. When x=5m, the acceleration is 42 m/s^2.
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..

Where are you getting a "y"?

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

Last edited:
Feldoh said:
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?

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

## 1. What is the formula for acceleration?

The formula for acceleration is a = (v2 - v1) / t, where a is acceleration, v2 is final velocity, v1 is initial velocity, and t is time.

## 2. How does the given velocity equation relate to acceleration?

The given velocity equation, v = x^3-4x^2+7x, is a function that represents the rate of change of an object's position. To determine acceleration, we need to find the second derivative of this function, which will give us the rate of change of velocity over time.

## 3. How can we use the given equation to determine acceleration?

To determine acceleration, we can differentiate the velocity equation twice. The first derivative will give us the rate of change of velocity, and the second derivative will give us the rate of change of acceleration. We can then substitute the given values of x into the second derivative to find the acceleration at a specific point.

## 4. Can the given equation be used to find acceleration at any given time?

Yes, the given equation can be used to find acceleration at any given time. We just need to plug in the value of time (t) into the second derivative of the velocity equation to get the acceleration at that specific time.

## 5. How can we interpret the acceleration value obtained from the given equation?

The acceleration value obtained from the given equation represents the rate of change of an object's velocity at a specific time. A positive acceleration value indicates that the object is accelerating in the positive direction, while a negative acceleration value indicates that the object is accelerating in the negative direction. A zero acceleration value means that the object's velocity is constant.

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