Average acceleration with different accelerations

  • Thread starter kylera
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  • #1
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Homework Statement


A motorboat starting from a dead stop accelerates at an average rate of 2.0 m/s^2 for 3 seconds, then very rapidly roars up to 4.0m/s^2 for 4 seconds. Finally, it decelerates to a stop 20 seconds later at a rate of 1.1m/s^2.
(a) What is its average acceleration during the first 27 seconds?
(b) What is its instantaneous acceleration 10 seconds into the trip?


Homework Equations


average acceleration = (v(f)-v(i)) / (f - i)
v = v(0) + at

The Attempt at a Solution


(a)

From t=0.0s to 3.0s,
v(0) = 0m/s
v(3) = 0 + 2.0 * 3 = 6.0m/s

From t=3.0s to 7.0s,
v(0) = v(3) ==> due to re-setting the frame of reference so t=3.0s => t=0.0s
v(4) = v(7) = 6.0 + 4.0*4 = 22m/s

From t=7.0s to 27.0s
v(final) = v(27) = 0.0m/s since it comes to a stop
average acceleration from t=7.0s to t=27.0s = -1.1m/s^2
Applying the average acceleration formula, -1.1 = (0 - V(i)) / 20, V(i) = v(7) = 22m/s

By following the average acceleration formula, I would all too easily get zero as average since V(i) and V(f) are both zero. What am I not seeing here?
 

Answers and Replies

  • #2
HallsofIvy
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Homework Statement


A motorboat starting from a dead stop accelerates at an average rate of 2.0 m/s^2 for 3 seconds, then very rapidly roars up to 4.0m/s^2 for 4 seconds. Finally, it decelerates to a stop 20 seconds later at a rate of 1.1m/s^2.
(a) What is its average acceleration during the first 27 seconds?
(b) What is its instantaneous acceleration 10 seconds into the trip?


Homework Equations


average acceleration = (v(f)-v(i)) / (f - i)
v = v(0) + at

The Attempt at a Solution


(a)

From t=0.0s to 3.0s,
v(0) = 0m/s
v(3) = 0 + 2.0 * 3 = 6.0m/s

From t=3.0s to 7.0s,
v(0) = v(3) ==> due to re-setting the frame of reference so t=3.0s => t=0.0s
v(4) = v(7) = 6.0 + 4.0*4 = 22m/s

From t=7.0s to 27.0s
v(final) = v(27) = 0.0m/s since it comes to a stop
average acceleration from t=7.0s to t=27.0s = -1.1m/s^2
Applying the average acceleration formula, -1.1 = (0 - V(i)) / 20, V(i) = v(7) = 22m/s

By following the average acceleration formula, I would all too easily get zero as average since V(i) and V(f) are both zero. What am I not seeing here?
Why all too easily? You don't like easy problems? Yes, if the object is moving at the same velocity (0) at both beginning and end, the average velocity is 0.
 
  • #3
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Oh.

I do like easy problems...it just seemed too easy that something seemed off.
 

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