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

AI Thread Summary
To find the net force on a training helicopter, the second derivative of its position vector must be taken to determine acceleration. The position vector is given as a function of time, and acceleration is derived from the velocity, which is the first derivative of position. Understanding this relationship is crucial since net force is calculated using Newton's second law, F = ma, where 'm' is mass and 'a' is acceleration. The discussion emphasizes the importance of these derivatives in solving for net force at a specific time, in this case, t=5.0 seconds. Clarification on the necessity of the second derivative helps solidify the connection between position, velocity, acceleration, and force.
Mdhiggenz
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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!
 
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Mdhiggenz said:

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.
 
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?
 
Mdhiggenz said:
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.
 
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