The position of a 2.75x10^5 N training helicopter under test is given by

[Mdhiggenz]

Homework Statement

The position of a 2.75x10^5 N training helicopter under test is given by...?
vector r = (0.020m/s^3)t^3ihat+(2.2 m/s)tjhat-(0.060m/s^2)t^2 khat

Find the net force on the helicopter at t=5.0s

Express your answer in terms of ihat, jhat , and khat . Express your coefficient using two significant figures.

Homework Equations

The Attempt at a Solution

I don't understand why me must take the second derivative, already solved it just want some clarification on to why we get the second derivative.

Thank you!

 

[cepheid]

Homework Statement

The position of a 2.75x10^5 N training helicopter under test is given by...?
vector r = (0.020m/s^3)t^3ihat+(2.2 m/s)tjhat-(0.060m/s^2)t^2 khat

Find the net force on the helicopter at t=5.0s

Express your answer in terms of ihat, jhat , and khat . Express your coefficient using two significant figures.

Homework Equations

The Attempt at a Solution

I don't understand why me must take the second derivative, already solved it just want some clarification on to why we get the second derivative.

Thank you!

By definition, the second derivative of position with respect to time is the acceleration. This should be pretty intuitive:

velocity is the rate of change of position with time.

acceleration is the rate of change of velocity with time.

Hence, velocity is the derivative of position, and acceleration is the derivative of velocity.

 

[Mdhiggenz]

I think I understand, since force = sigmaF = Ma

we have the force, and we are solving for the acceleration in order to multiply it by the mass to get the net?

 

[cepheid]

I think I understand, since force = sigmaF = Ma

we have the force, and we are solving for the acceleration in order to multiply it by the mass to get the net?

Yes, exactly. If you know the acceleration, and the mass, then you know the net force. That is because of Newton's 2nd Law.

 

[asteriatic]

I would like to clarify that the second derivative is necessary in this case because the given position vector is a function of time, and the net force acting on an object is equal to its mass multiplied by its acceleration, which is the second derivative of its position with respect to time. Therefore, in order to find the net force on the helicopter at t=5.0s, we need to take the second derivative of the given position vector with respect to time and then multiply it by the mass of the helicopter. This will give us the net force vector in terms of ihat, jhat, and khat with a coefficient of two significant figures. I hope this helps to clarify the reasoning behind taking the second derivative in this situation.