Kinematics problem-finding acceleration

Click For Summary
Mary and Sally are in a foot race where Mary needs to determine the constant acceleration required to finish side-by-side with Sally. Sally, initially faster, decelerates at 0.48 m/s², and calculations show that Mary must cover 22 meters while Sally covers 17 meters. The initial calculations led to an incorrect acceleration of 0.60 m/s² due to rounding errors and miscalculating the distance Mary needed to run. After correcting these mistakes, the required acceleration for Mary was found to be approximately 0.53 m/s². Attention to detail in calculations and avoiding premature rounding were emphasized as critical for accurate results.
life is maths
Messages
37
Reaction score
0

Homework Statement


Hi everyone! This is my homework question, an easy one, but I cannot find my mistake. Any help will be greatly appreciated.

Mary and Sally are in a foot race. When Mary is 22 m/s from the finish line, she has a speed of 4.0 m/s and is 5.0 m behind Sally, who has a speed of 5.0 m/s. During the remaining portion of the race, Sally decelerates at a constant rate of 0.48 m/s2 to the finish line.

What constant acceleration does Mary now need during the remaining portion of the race, if she wishes to cross the finish line side-by-side with Sally?

Homework Equations





The Attempt at a Solution


(The final velocity of Sally)=Vsf2
(The initial velocity of Sally)=Vsi2
Vsf2:Vsi2-2ax
Vsf2=52-2*0.48*17
which is 3 m/s.
Vsf=Vsi+at
3=5-0.48t
t=4.16s

They should finish the race at the same time, so this is also the time in which Mary ran 17 m.

X=Vot+1/2at2
22=4*4.16+a*(4.16)2
5=2.08*4.16*a
and a=0.60 m/s2

This is an internet assignment and the program says this result is false.
Is there a mistake here? What do I do wrong? Thanks for any help :)
 
Physics news on Phys.org
life is maths said:

Homework Statement


Hi everyone! This is my homework question, an easy one, but I cannot find my mistake. Any help will be greatly appreciated.

Mary and Sally are in a foot race. When Mary is 22 m/s from the finish line, she has a speed of 4.0 m/s and is 5.0 m behind Sally, who has a speed of 5.0 m/s. During the remaining portion of the race, Sally decelerates at a constant rate of 0.48 m/s2 to the finish line.

What constant acceleration does Mary now need during the remaining portion of the race, if she wishes to cross the finish line side-by-side with Sally?

Homework Equations


The Attempt at a Solution


(The final velocity of Sally)=Vsf2
(The initial velocity of Sally)=Vsi2
Vsf2:Vsi2-2ax
Vsf2=52-2*0.48*17
which is 3 m/s.
Vsf=Vsi+at
3=5-0.48t
t=4.16s

They should finish the race at the same time, so this is also the time in which Mary ran 17 m.

X=Vot+1/2at2
22=4*4.16+a*(4.16)2
5=2.08*4.16*a
and a=0.60 m/s2

This is an internet assignment and the program says this result is false.
Is there a mistake here? What do I do wrong? Thanks for any help :)

Homework Statement


Homework Equations


The Attempt at a Solution


Your final velocity of 3 m/s is 2.946... rounded off.
Not sure you can round off like that during the calculations.

EDIT: and also Mary had to run 22m. ( she was 5m behind)
 
Be careful about rounding off your intermediate calculations. For example, don't round off Sally's final speed to 3 m/s--that introduces errors in all subsequent calculations.
 
I did not round off the velocity of Sally, and this time I have t=4.375 s. When I do calculations according to the new value of t, I find the acceleration of Mary as 0.47 m/s^2. I still cannot figure out my mistake, and I used my last chance to answer it. Anyways, this question is a pain in the neck for me for a week.

My mistake that Mary ran 17 m, but I didn't do my calculations according to this mistake.
Thanks for your help :)
 
life is maths said:
I did not round off the velocity of Sally, and this time I have t=4.375 s.
I got 4.279 s.
 
Doc Al said:
I got 4.279 s.

That's true! I have no idea how I could find 4.375 s.

Then

22=4t+1/2at2
22-17.116=9.154a
4.884=9.154a
a=0.53m/s2

I should have been more careful with my calculations and rounding off... I've lost 20 points already, but at least I could solve it. Thanks a lot for your help, Doc Al :)
 
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 1 ·
Replies
1
Views
2K
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
15K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K