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Acceleration, Velocity and Position (calculus)

  • Thread starter adoado
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  • #1
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Homework Statement



The acceleration of a particle is directly proportional to time, t. At t = 0, the velocity of that particle, v, = 16 in./sec. Knowing that v = 15 in/sec and x = 20in when t = 1, determine the velocity, acceleration and position at x = 7.

Homework Equations



The Attempt at a Solution



I took
[tex]\frac{dv}{dt} = kt, so \: v = \frac{k}{2}t^2 + C[/tex]

Solving for C:
[tex]16 = \frac{k}{2}(0)^2 + C[/tex]
C = 16

Solve for k

[tex]1 = \frac{k}{2}(1)^2 + 16[/tex]

k = -30

Therefore:

[tex]v(t) = \-15*(t)^2 + 16[/tex]

This is already wrong; v(7) = -719 in/sec but the book states the answer is -33 in/sec

Any advice on what's gone wrong would be greatly appreciated; this is a revision question too...

Cheers,
Adrian
 

Answers and Replies

  • #2
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k is -2
 
  • #3
ehild
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You need to find the velocity when x = 7. Find x(t) first.

ehild
 
  • #4
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Ah, thankyou ehild; I failed to read the question once again... I will give it a shot. How is k = -2?
 
  • #5
ehild
Homework Helper
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Ah, thankyou ehild; I failed to read the question once again... I will give it a shot. How is k = -2?
Looks OK.
 
  • #6
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I found x(t) by integrating and finding C. I obtained this:

[tex]x(t)=16t-5t^3+9[/tex]

Then solving for t when x = 7

[tex]16t-5t^3+2 = 0[/tex]

Sorry, how does one solve this?

Cheers,
Adrian
 
Last edited:
  • #7
ehild
Homework Helper
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Don't forget that k=-2, so v(t)=16-t^2.

Looking at the solution for v(t) at x=7 , it should be rather t=7 instead of x. Otherwise why would they ask the position at x=7?

ehild
 
Last edited:
  • #8
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Sorry, this is probably really stupid, but where/how does k = -2?
 
  • #9
ehild
Homework Helper
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"v = 15 in/sec and x = 20in when t = 1"

15=k/2*1^2+16

ehild
 
  • #10
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Ah! I put in a bad number...dodgy mistake indeed; Cheers, I will retry!
 
  • #11
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Got it!! Thanks so much, both of you! ;)
 

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