Average acceleration with different accelerations

But I now understand that the object's velocity is 0 at the beginning and end, and therefore the average velocity is also 0.
  • #1
kylera
40
0

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?
 
Physics news on Phys.org
  • #2
kylera said:

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
Oh.

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

Related to Average acceleration with different accelerations

What is average acceleration with different accelerations?

Average acceleration with different accelerations refers to the calculation of the overall rate of change in velocity over a given time period when an object experiences varying accelerations. It takes into account the initial and final velocities, as well as the time elapsed during which the accelerations occur.

How is average acceleration with different accelerations calculated?

To calculate the average acceleration with different accelerations, you can use the formula: average acceleration = (final velocity - initial velocity) / time elapsed. This formula can be applied for each individual acceleration and then the average of all the values can be taken to get the overall average acceleration.

What are some real-life examples of average acceleration with different accelerations?

Some real-life examples of average acceleration with different accelerations include a car accelerating from a standstill to a certain speed while encountering traffic and stoplights, a rollercoaster accelerating and decelerating at different points along the track, and a rocket launching into space with varying levels of thrust.

How does average acceleration with different accelerations relate to Newton's Second Law of Motion?

Newton's Second Law of Motion states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass. Average acceleration with different accelerations takes into account the net force acting on an object and the time period over which it is applied, making it a practical application of this law.

What are the limitations of using average acceleration with different accelerations?

One limitation of using average acceleration with different accelerations is that it assumes a constant rate of change in acceleration over the given time period. In real-life scenarios, accelerations may vary continuously, making it difficult to accurately calculate the average. Additionally, it does not take into account factors such as air resistance or friction, which can affect the overall acceleration of an object.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
474
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
2
Replies
42
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
899
  • Introductory Physics Homework Help
2
Replies
38
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
1K
Back
Top