MHB Mechanics- General motion in a straight line.

AI Thread Summary
A woman on a sledge moves across ice with an initial velocity of 2 m/s and an acceleration of a = -0.01t m/s². Calculations show that she travels approximately 26.7 meters before coming to rest, based on integrating the velocity and distance equations. There is disagreement regarding the textbook answer of 8.43 meters, which may stem from a potential typo in the acceleration value. The discussion concludes that if the acceleration were indeed -0.1t m/s², the textbook answer would be correct, but with the given acceleration, 26.7 meters is the accurate distance. The resolution emphasizes the importance of precise values in physics calculations.
Shah 72
MHB
Messages
274
Reaction score
0
A woman on a sledge moves in a straight line across horizontal ice. Her initial velocity is 2 m/s. Throughout the journey her acceleration is given by a= -0.01t m/s^2, where t is the time from the start in seconds. Find the distance that she travels before coming to rest.
Iam getting the ans 53.3 m
When t= 0s initial velocity= 2m/s
I integrated to get v= -0.01t^2/2+2
I integrated v to get s= -0.01t^3/6 +2t+c
t=0, s=0, c=0
When it comes to rest v= 0
I get t=20s taking the velocity equation.
I substitute in s and got 53.3m
The ans in textbook is 8.43m
 
Mathematics news on Phys.org
I disagree with both your solution and the text “answer”.

$\Delta x = 2t - \dfrac{0.01t^3}{6} = 40 - \dfrac{80}{6} = \dfrac{80}{3} \approx 26.7$ m

which matches up with the evaluation of the definite integral

$\displaystyle \Delta x = \int_0^{20} 2 - \dfrac{0.01t^2}{2} \, dt$

if the acceleration was $-0.1t \, m/s^2$, then the text solution is correct
 
skeeter said:
I disagree with both your solution and the text “answer”.

$\Delta x = 2t - \dfrac{0.01t^3}{6} = 40 - \dfrac{80}{6} = \dfrac{80}{3} \approx 26.7$ m

which matches up with the evaluation of the definite integral

$\displaystyle \Delta x = \int_0^{20} 2 - \dfrac{0.01t^2}{2} \, dt$

if the acceleration was $-0.1t \, m/s^2$, then the text solution is correct
Sorry that was a typo mistake. I got 26.7 m.
So the textbook ans is wrong.
Thank you so so much!
 
Shah 72 said:
Sorry that was a typo mistake. I got 26.7 m.
So the textbook ans is wrong.
Thank you so so much!

As I stated, the text solution was not incorrect if it was a typo with the acceleration by one decimal place … which is a plausible explanation imo.
 
skeeter said:
I disagree with both your solution and the text “answer”.

$\Delta x = 2t - \dfrac{0.01t^3}{6} = 40 - \dfrac{80}{6} = \dfrac{80}{3} \approx 26.7$ m

which matches up with the evaluation of the definite integral

$\displaystyle \Delta x = \int_0^{20} 2 - \dfrac{0.01t^2}{2} \, dt$

if the acceleration was $-0.1t \, m/s^2$, then the text solution is correct
Yeah I got that. Textbook question has a=- 0.01t m/s^2
So according to this our ans of 26.7m should be right. And as you mentioned if the question had a= -0.1t m/ s^2 then s= 8.43 m.
Thanks a lot!
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...

Similar threads

Replies
2
Views
2K
Replies
6
Views
1K
Replies
4
Views
987
Replies
7
Views
922
Replies
2
Views
981
Replies
5
Views
1K
Replies
9
Views
1K
Back
Top