Acceleration with limited known variables Question

In summary, the bush baby is capable of leaping vertically to a height of 2.26 m by accelerating over a distance of 0.160 m. The acceleration during the jump is constant and can be calculated using the equation a = Vo²/(2y), where Vo is the initial velocity and y is the distance traveled. The calculated acceleration is 128.625 m/s^2 or approximately 13.125 g's. However, the correct answer is 138 m/s^2, indicating a potential error in the calculation or a different method used to obtain the answer.
  • #1
bbauer2
5
0

Homework Statement



A bush baby, an African primate, is capable of leaping vertically to the remarkable height of 2.26 m. To jump this high, the bush baby accelerates over a distance of 0.160 m, while extending the legs. The acceleration during the jump is approximately constant. What is the acceleration in m/s?

h=2.26m
d=.160m
g=9.8m/s

Homework Equations



d=.5 X a X t
V=a X t
h=V X t-1/2g(t^2)
0=V - g X t


The Attempt at a Solution



I have no idea how to attempt this without another variable given.
 
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  • #2
bbauer2 said:

Homework Statement



A bush baby, an African primate, is capable of leaping vertically to the remarkable height of 2.26 m. To jump this high, the bush baby accelerates over a distance of 0.160 m, while extending the legs. The acceleration during the jump is approximately constant. What is the acceleration in m/s?

h=2.26m
d=.160m
g=9.8m/s

Homework Equations



d=.5 X a X t
V=a X t
h=V X t-1/2g(t^2)
0=V - g X t


The Attempt at a Solution



I have no idea how to attempt this without another variable given.

Acceleration is in m/s^2, not m/s.

The acceleration happens first, to give a Vo. Then the rest is just governed by the simple/regular gravity kinematic equations. Show your work.
 
  • #3
Here's what I got, and I submitted it and it's still wrong. Can anyone tell me where I am going wrong?

The square of the max initial velocity Vo² = 2gh where h is 2.16 - .16 = 2.1 m.

Vo² = 41.16 m²/s²

To reach this speed in a jump of y meters, a = Vo²/(2y) = 41.16/(2*.16)

a = 128.625 m/s = 13.125 g's
 
  • #4
bbauer2 said:
Here's what I got, and I submitted it and it's still wrong. Can anyone tell me where I am going wrong?

The square of the max initial velocity Vo² = 2gh where h is 2.16 - .16 = 2.1 m.

Vo² = 41.16 m²/s²

To reach this speed in a jump of y meters, a = Vo²/(2y) = 41.16/(2*.16)

a = 128.625 m/s = 13.125 g's

The numbers look right to me, except for the units on the final answer (I'll say again, the units of acceleration are m/s^2, not m/s). If you carried the units along in your equations above the answer, you would have gotten the same number and the correct units.

If you submit a = 128.63 m/s^2, is it still wrong?
 
  • #5
Yeah, it's wrong. All I need is the numbers, the m/s^2 is given. It said the answer is 138! Anyone know how they got that?
 

Related to Acceleration with limited known variables Question

1. What is acceleration with limited known variables?

Acceleration with limited known variables is a scientific concept that describes the rate at which an object changes its velocity over time, when the initial and final velocities, as well as the time interval, are not fully known.

2. How is acceleration with limited known variables calculated?

Acceleration with limited known variables can be calculated using the formula a = (vf - vi)/t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time interval.

3. What are some examples of limited known variables in acceleration?

Examples of limited known variables in acceleration include objects changing direction or speed without a known initial or final velocity, or acceleration caused by multiple forces acting on an object.

4. How does acceleration with limited known variables differ from normal acceleration?

The main difference between acceleration with limited known variables and normal acceleration is that in the former, the initial and final velocities, as well as the time interval, are not fully known, making it more challenging to calculate the acceleration accurately.

5. Why is it important to understand acceleration with limited known variables?

Understanding acceleration with limited known variables is crucial in many scientific fields, such as physics and engineering, as it allows for the accurate prediction and analysis of the motion of objects in real-world scenarios, where not all variables are always known or controllable.

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