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Physics Mechanics- General motion in a straight line.

Shah 72

Member
Apr 14, 2021
193
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
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
935
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
 

Shah 72

Member
Apr 14, 2021
193
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!!!
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
935
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.
 

Shah 72

Member
Apr 14, 2021
193
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!