Instantaneous and average velocity help

  • #1
vAn objects moves horizontally according to the following equation s(t)=t3-2t2-5t+6. Determine the instantaneous velocity of the object at t0=1 second using the definition of a derivative at a point.

v'(t)= lim (s(t0+h)-s(t0))/h
h-->0


And
What is the average velocity of the object from t=0 to t=2?
Average velocity= (s(t2)-s(t1))/(t2-t1)



For the first one, i got:
((1+h)3-2(1+h)2-5(1+h)+6-13-2(1)2-5(1)+6)/h

I expanded and factored all of that ang got (h(-1+h+h2)/h then i plugged in 0 for h and got -1 as the instantaneous velocity... is that correct?

I got the first one... i just added the values wrong... i got -6. But i still don't know how to do the second one.

and for the second one, i have (3(2)2-4(2)-5)/2-0 but that makes it 0/2 and i know that this can't be.

Can you please help me?thank you!
 
Last edited:

Answers and Replies

  • #2
For the second part, it looks like you used the derivative of s (i.e. the velocity) in the numerator. You should use s itself (i.e. plug t1 and t2 into s(t) for the numerator) as in the definition of average velocity.
 
  • #3
For the second part, it looks like you used the derivative of s (i.e. the velocity) in the numerator. You should use s itself (i.e. plug t1 and t2 into s(t) for the numerator) as in the definition of average velocity.

so it should just be (-4-0)/(2-0)=-1/2?
 
  • #4
I think so. No. Wait a minute. Check it again. I think your t1 part is wrong. Just check it over carefully.
 
  • #5
wait wouldn't it be -5 for the average velocity instead?
 
  • #6
Yeah. That's what I get.
 
  • #7
thank you!
 

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