1-D (I think) Speed to distance problem

  • Thread starter Thread starter magnanimousto
  • Start date Start date
  • Tags Tags
    Speed
Click For Summary
A runner accelerates from rest to 2.0 m/s over 12 m, and the problem asks for the distance needed to reach 3.0 m/s under the same force. The initial attempt to solve it using time was incorrect, leading to confusion. Participants suggest using kinematic equations or the Work-Energy Theorem to find acceleration first. Recognizing that the force remains constant implies the acceleration is the same for both scenarios is crucial. The correct total distance to accelerate to 3.0 m/s is determined to be 27 m.
magnanimousto
Messages
10
Reaction score
0

Homework Statement



A runner exerts a constant force to accelerate from rest to 2.0 m/s over a distance of 12 m.
Assuming the runner can keep up the same force, what total distance would be needed to ac-
celerate up from rest to 3.0 m/s?

Homework Equations



none given.

The Attempt at a Solution



The answer to the problem is 27. After quite some time spent thinking I tried solving first part for time using x= 1/2 vt and then plugging it into 2nd part. but that gives time as 12 and 2nd x as 18 so wrong. Dunno how to solve it
 
Physics news on Phys.org
magnanimousto said:

Homework Statement



A runner exerts a constant force to accelerate from rest to 2.0 m/s over a distance of 12 m.
Assuming the runner can keep up the same force, what total distance would be needed to ac-
celerate up from rest to 3.0 m/s?

Homework Equations



none given.

The Attempt at a Solution



The answer to the problem is 27. After quite some time spent thinking I tried solving first part for time using x= 1/2 vt and then plugging it into 2nd part. but that gives time as 12 and 2nd x as 18 so wrong. Dunno how to solve it
Hello magnanimousto. Welcome to PF!

What kinematic equations do you know?

Alternatively, do you know the Work-Energy Theorem ?
 
This is a constant acceleration problem, so you can use the basic kinematic equations for that case. Try solving for the acceleration first, and using that value for the next part.

If you don't know those equations, here's the one you should probably use -- they're very helpful to memorize though.
v_{f}^{2}= v_{i}^{2} + 2aΔd
 
Thanks jackarms & SammyS.I do know kinematics equations, for some reason it never occurred to me that since the force is the same in both so must be the acceleration
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 8 ·
Replies
8
Views
830
Replies
20
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
1
Views
2K
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
8K
Replies
24
Views
3K
Replies
4
Views
2K