One-Dimensional Kinematics Problem

In summary, Jackie is having trouble with two questions on the Physics Cappedstone. She correctly calculates how fast she is when she reaches the ground, but incorrectly uses the general case when calculating the deceleration. She also correctly calculated how much work she does when she jumps, but is having trouble with the final equation.
  • #1
UCrazyBeautifulU
33
0
Hey guys,

There is one more question I am having a problem with. Here it is:

You jump from the top of a boulder to the ground 2.07 m below. Calculate your deceleration on landing. Assume that in order to soften your landing, your legs will bend 0.510 m. Give your answer in units of g. Do not enter unit.

Here is what I have done so far. I have used the height to find out how fast you are when you reach the ground. I got v= 6.37 m/s. After that I know I have to use the distance of bending to determine over what distance you decelerate. I know I use the formula that relates the velocity to the distance. So I should use v^2 = v_o^2 - 2g(xf-xi). THat is the same formula I used to get the first answer. Maybe I am using the wrong formula. This problem and the one I posted before I have made like 10 attempts on LON CAPA and I don't know what to do anymore. Please help. Thank you,

jackie

You can also send me an email if you don't want to post on the board.
 
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  • #2
You correctly calculated how fast you are when you reach the ground:
v2 = 2gh

Relating work, Newton 2nd Law and Work-Energy Theorem we have:
W = max = \Delta KE = 0 - 0.5mv2
max = 0 - 0.5mv2
max = -0.5m2gh
max = -mgh
 
  • #3
I don't know any of those formulas or anything you listed above. I just started physics so we haven't gotten to those yet. Thanks for letting me know I got the first part right. I need things in simple terms I guess.
 
  • #4
UCrazyBeautifulU said:
So I should use v^2 = v_o^2 - 2g(xf-xi).

That equation is almost correct, you should however be using the general case (not just for objects falling under gravity) thus;

[tex]v^2 = v_{0}^{2} + 2a(x_{f} - x_{i})[/tex]

Where a is acceleration. Don't forget to divide your answer by 'g' to obtain your answer in terms of g.
 
  • #5
so v^2 is 40.61, what should I put in for v_o? and what should I put in for the change in x in that equation. If i put in 0 as initial velocity and change in x as 0.510 i end up with acceleration -39.81, divide that by 9.81 and i get -4.06 which was incorrect according to lon capa.
 
  • #6
UCrazyBeautifulU said:
so v^2 is 40.61, what should I put in for v_o? and what should I put in for the change in x in that equation. If i put in 0 as initial velocity and change in x as 0.510 i end up with acceleration -39.81, divide that by 9.81 and i get -4.06 which was incorrect according to lon capa.
Note quite, v2 represents the final velocity, which in this case is zero. V0 would be the velocity at which he reaches the ground. Do you follow?
 
  • #7
even if i switch the two v's i end up with the same answer just different sign, unless i am not supposed to put in 0.510 for x. I am ready to quit physics.
 
  • #8
Well, I've quickly checked your calculations and I can find nothing wrong. Have you tried entering the modulus of the acceleration (without the negative sign). Make sure your not entering any units as the value has no units. Are you using their quoted value of g?
 
  • #9
Hey, thank you so much for that advice Hoot. I was putting in -4.06 and i tried it without the negative and it worked. THANK YOU. Can you look at my other problem I posted ? please!
 

What is one-dimensional kinematics?

One-dimensional kinematics is the study of motion in a single direction without considering the causes of the motion.

What are the basic concepts of one-dimensional kinematics?

The basic concepts of one-dimensional kinematics include displacement, velocity, acceleration, and time.

How is displacement different from distance?

Displacement is the change in position of an object from its initial position to its final position, while distance is the total length traveled by an object regardless of its direction.

What is the difference between average velocity and instantaneous velocity?

Average velocity is the total displacement divided by the total time, while instantaneous velocity is the velocity at a specific moment in time.

How is acceleration calculated in one-dimensional kinematics?

Acceleration is calculated by dividing the change in velocity by the change in time, or by taking the second derivative of the position function with respect to time.

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