derivative frq type problem

[brazen needle]

For 3. the original function gives you the position and so the derivative is change in position (distance) over change in time a.k.a velocity. So the derivative defines the velocity of the particle as a function of time. So when is that function negative? For 5. you want to find an expression for acceleration. Keep in mind acceleration is change in velocity over change in time, a.k.a the slope of the velocity function. Can you find an expression for this given a function for velocity? Then once you've found the function for acceleration as a function of time, find the values of time where the acceleration is zero and check whether these values fulfill that the particle is moving (it's moving when velocity isn't 0)

[dusky helm]

I’m so sorry I didn’t know how to close this channel. But I appreciate the entire paragraph you wrote trying to help 😅

[brazen needle]

Oh so you've solved it?

[dusky helm]

Yes on my own

[brazen needle]

Alright no worries

[dusky helm]

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