The position function of a particle is given by

s=t^3-4.5t^2-7t or equal to 0.

When does the particle reach a velocity of 5 m/s?

Question

The position function of a particle is given by

s=t^3-4.5t^2-7t > or equal to 0.

When does the particle reach a velocity of 5 m/s?

anonymous · Answer

So I took the derivative of this and then plugged 5 in for t, and I got 23 seconds. The answer in my book says 4 seconds, though.

kainui · Answer

You plugged in 5m/s for t, but t is time not velocity!

anonymous · Answer

Oh! I see, so how would I solve this?

kainui · Answer

Well, in the equation you started with, what is "s"?

When you take the derivative you get s', or $$s'=\frac{ds}{dt}$$So what's that?

anonymous · Answer

the derivative of s is 3t^2-9t-7

kainui · Answer

Yes, but that's not what I mean.

For instance I can ask you what pi is, and you can say 3.1415... etc... But that's not answering the question. That's just the _value_ pi has. Pi is the ratio of a circle's circumference to its diameter.

Similarly here, you're showing me what value the derivative of s has, but what does it mean?

anonymous · Answer

Would it be: the position of a particle?

kainui · Answer

s is the position, but the derivative of s is the change of s. So just replace "s" with "position" and you can see that the derivative of position is the change of position. What's the change of position? The velocity.

So the derivative of position is velocity since it's just the change of position with respect to time. Does that make sense? If you understand this it'll make physics a lot simpler to understand.

anonymous · Answer

Yeah, I think I understand! So would the derivative of s be the velocity of that particle?

kainui · Answer

Exactly! So now does it make sense where to plug in the 5 m/s in? Then solving for time should work out. =)

anonymous · Answer

Oh, would it be 5= 3t^2-9t-7?

kainui · Answer

Perfect. So now you have a quadratic equation to solve. This means you might get two answers, however pay careful attention to what the question is asking. Usually things start at t=0 so negative times can be thrown out. In this case it sounds like it's asking when does it first reach 5 m/s, so it might speed up then slow down, just as a reminder for future problems, and that's OK, just remember to think about what's physically happening.

anonymous · Answer

Wow, great! Thank you! :)

kainui · Answer

Glad I could help, a kind of cool way to think of it is when you were younger you probably learned distance is rate times time, d=rt. But now if we rephrase this in terms of these, 

s is position, or distance
s' is velocity, or rate
t is still time

So now $$s=\frac{ds}{dt}t$$ is just sort of one case of using derivatives.