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The position of an object is given by $${x(t)= 2.17+4.80t^2-0.100t^6}$$

Would I have to take the derivative of that in order to find velocity?

But from there how would I find the position and acceleration the instant velocity is zero? For time I thought about setting v to zero and solving to find the time but I don't know how that would really help me.

Would acceleration just be the derivative of v? Is there an exact way to find acceleration?

Andrea
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  • "For time I thought about setting v to zero and solving to find the time but I don't know how that would really help me." Do it anyway! :) – Andrew Dudzik Feb 09 '15 at 01:52
  • I did and one was 0 sec and the other was 2 sec. I figured 2 sec. would be more helpful right? But I don't know which equation would help me find acceleration and position – Andrea Feb 09 '15 at 01:53
  • Your first sentence is "the position of an object is given by" so you should have no trouble finding the position with your own equation. For acceleration, you correctly guessed how to find the equation at the end of your edited post. – Andrew Dudzik Feb 09 '15 at 01:58
  • By the way, the question as stated is ambiguous: As you have discovered, there are two moments when the velocity is zero, $t=0$ and $t=2$. These will give you two different results for the position and acceleration. – Andrew Dudzik Feb 09 '15 at 02:00

1 Answers1

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The position of an object is given by

$${x(t)= 2.17+4.80t^2-0.100t^6}$$

Thus velocity of the object is given by x'(t) which is:

$$x'(t)=v(t) = 9.60t-0.600t^5$$

Since velocity is 0, we can solve for time:

$$0= 9.60t-0.600t^5$$

$$ t= -2,0,2$$

Since time can't be negative, and we probably want the positive t -value, let's use t = 2.

We then sub t = 2 into x(t) and a(t).

We can get a(t) by using the identity v'(t)=a(t)

$$v'(t)=a(t)=9.6-3t^4$$

Solving for t =2:

$$a(2)=9.6-3*2^4=-38.4$$

$${x(2)= 2.17+4.80*2^2-0.100*2^6=14.97}$$

Therefore, position and acceleration when velocity is equal to 0, is 14.97 $units$ and -39.4 $units/s^2$

Victor
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