MHB Mechanics- General motion in a straight line.

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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
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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
 
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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!
 
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