How to find speed with friction?
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You testify as an expert witness in a case involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill. You find that the slope of the hill is θ = 14.5°, that the cars were separated by distance d = 22.0 m when the driver of car A put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car A at the onset of braking was v0 = 18.5 m/s.
(a) With what speed did car A hit car B if the coefficient of kinetic friction was 0.60 (dry road surface)?
m/s
(b) What was the speed if the coefficient of kinetic friction was 0.10 (road surface covered with wet leaves)?
m/s
How can this be solved? The book says , a = g sin angle - coefficient cos angle = a negative numver. But when I try it, the number is positive...
physics
asked Mar 7 '17 at 6:35
user416503user416503
62115
$endgroup$
add a comment |
$begingroup$
You testify as an expert witness in a case involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill. You find that the slope of the hill is θ = 14.5°, that the cars were separated by distance d = 22.0 m when the driver of car A put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car A at the onset of braking was v0 = 18.5 m/s.
(a) With what speed did car A hit car B if the coefficient of kinetic friction was 0.60 (dry road surface)?
m/s
(b) What was the speed if the coefficient of kinetic friction was 0.10 (road surface covered with wet leaves)?
m/s
How can this be solved? The book says , a = g sin angle - coefficient cos angle = a negative numver. But when I try it, the number is positive...
physics
asked Mar 7 '17 at 6:35
user416503user416503
62115
$endgroup$
$begingroup$
As an expert witness you're probably supposed to base this on actual physics as opposed to naive mathematical models. This therefore becomes more of a physics question. Note for example that the friction for a car is not modelled using coefficient of friction.
$endgroup$
– skyking
Mar 7 '17 at 7:53
$begingroup$
@skyking This is obviously an artificial word problem that involves a fictional and irrelevant expert witness for the story.
$endgroup$
– amd
Mar 7 '17 at 8:39
add a comment |
$begingroup$
You testify as an expert witness in a case involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill. You find that the slope of the hill is θ = 14.5°, that the cars were separated by distance d = 22.0 m when the driver of car A put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car A at the onset of braking was v0 = 18.5 m/s.
(a) With what speed did car A hit car B if the coefficient of kinetic friction was 0.60 (dry road surface)?
m/s
(b) What was the speed if the coefficient of kinetic friction was 0.10 (road surface covered with wet leaves)?
m/s
How can this be solved? The book says , a = g sin angle - coefficient cos angle = a negative numver. But when I try it, the number is positive...
physics
asked Mar 7 '17 at 6:35
user416503user416503
62115
$endgroup$
You testify as an expert witness in a case involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill. You find that the slope of the hill is θ = 14.5°, that the cars were separated by distance d = 22.0 m when the driver of car A put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car A at the onset of braking was v0 = 18.5 m/s.
(a) With what speed did car A hit car B if the coefficient of kinetic friction was 0.60 (dry road surface)?
m/s
(b) What was the speed if the coefficient of kinetic friction was 0.10 (road surface covered with wet leaves)?
m/s
How can this be solved? The book says , a = g sin angle - coefficient cos angle = a negative numver. But when I try it, the number is positive...
physics
physics
asked Mar 7 '17 at 6:35
user416503user416503
62115
asked Mar 7 '17 at 6:35
user416503user416503
62115
asked Mar 7 '17 at 6:35
user416503user416503
62115
asked Mar 7 '17 at 6:35
user416503user416503
62115
asked Mar 7 '17 at 6:35
user416503user416503
62115
62115
$begingroup$
As an expert witness you're probably supposed to base this on actual physics as opposed to naive mathematical models. This therefore becomes more of a physics question. Note for example that the friction for a car is not modelled using coefficient of friction.
$endgroup$
– skyking
Mar 7 '17 at 7:53
$begingroup$
@skyking This is obviously an artificial word problem that involves a fictional and irrelevant expert witness for the story.
$endgroup$
– amd
Mar 7 '17 at 8:39
add a comment |
$begingroup$
As an expert witness you're probably supposed to base this on actual physics as opposed to naive mathematical models. This therefore becomes more of a physics question. Note for example that the friction for a car is not modelled using coefficient of friction.
$endgroup$
– skyking
Mar 7 '17 at 7:53
$begingroup$
@skyking This is obviously an artificial word problem that involves a fictional and irrelevant expert witness for the story.
$endgroup$
– amd
Mar 7 '17 at 8:39
$begingroup$
As an expert witness you're probably supposed to base this on actual physics as opposed to naive mathematical models. This therefore becomes more of a physics question. Note for example that the friction for a car is not modelled using coefficient of friction.
$endgroup$
– skyking
Mar 7 '17 at 7:53
$begingroup$
As an expert witness you're probably supposed to base this on actual physics as opposed to naive mathematical models. This therefore becomes more of a physics question. Note for example that the friction for a car is not modelled using coefficient of friction.
$endgroup$
– skyking
Mar 7 '17 at 7:53
$begingroup$
@skyking This is obviously an artificial word problem that involves a fictional and irrelevant expert witness for the story.
$endgroup$
– amd
Mar 7 '17 at 8:39
$begingroup$
@skyking This is obviously an artificial word problem that involves a fictional and irrelevant expert witness for the story.
$endgroup$
– amd
Mar 7 '17 at 8:39
add a comment |
1 Answer
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oldest
votes
$begingroup$
Let the $x$ axis be along the road, downwards and let the $y$ axis be perpendicular to the road upwards. Projection of gravitational acceleration on the x-axis is $gsin{theta}$ and $-gcos{theta}$ on the y-axis. Friction is along the negative $x$ axis and is $F_{xf}=-gamma{v_{x}}$.
Newton along $x$ is
$$dot{v}_{x}=gsin{theta}-frac{gamma}{m}{v_{x}}$$
the solution is
$${v_{x}}(t)=frac{mg}{gamma}sin{theta}-frac{mc}{gamma}e^{-frac{gamma}{m}t}$$
At $t=0$ the speed was $v_{0x}$, so
$$c=gsin{theta}-frac{gamma}{m}v_{0x}$$
And the solution with initial conditions is
$${v_{x}}(t)=(1-e^{-frac{gamma}{m}t})frac{mg}{gamma}sin{theta}+v_{0x}e^{-frac{gamma}{m}t}$$
Use your numbers and calculate
answered Mar 7 '17 at 7:45
Kiryl PesotskiKiryl Pesotski
1,734410
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let the $x$ axis be along the road, downwards and let the $y$ axis be perpendicular to the road upwards. Projection of gravitational acceleration on the x-axis is $gsin{theta}$ and $-gcos{theta}$ on the y-axis. Friction is along the negative $x$ axis and is $F_{xf}=-gamma{v_{x}}$.
Newton along $x$ is
$$dot{v}_{x}=gsin{theta}-frac{gamma}{m}{v_{x}}$$
the solution is
$${v_{x}}(t)=frac{mg}{gamma}sin{theta}-frac{mc}{gamma}e^{-frac{gamma}{m}t}$$
At $t=0$ the speed was $v_{0x}$, so
$$c=gsin{theta}-frac{gamma}{m}v_{0x}$$
And the solution with initial conditions is
$${v_{x}}(t)=(1-e^{-frac{gamma}{m}t})frac{mg}{gamma}sin{theta}+v_{0x}e^{-frac{gamma}{m}t}$$
Use your numbers and calculate
answered Mar 7 '17 at 7:45
Kiryl PesotskiKiryl Pesotski
1,734410
$endgroup$
add a comment |
$begingroup$
Let the $x$ axis be along the road, downwards and let the $y$ axis be perpendicular to the road upwards. Projection of gravitational acceleration on the x-axis is $gsin{theta}$ and $-gcos{theta}$ on the y-axis. Friction is along the negative $x$ axis and is $F_{xf}=-gamma{v_{x}}$.
Newton along $x$ is
$$dot{v}_{x}=gsin{theta}-frac{gamma}{m}{v_{x}}$$
the solution is
$${v_{x}}(t)=frac{mg}{gamma}sin{theta}-frac{mc}{gamma}e^{-frac{gamma}{m}t}$$
At $t=0$ the speed was $v_{0x}$, so
$$c=gsin{theta}-frac{gamma}{m}v_{0x}$$
And the solution with initial conditions is
$${v_{x}}(t)=(1-e^{-frac{gamma}{m}t})frac{mg}{gamma}sin{theta}+v_{0x}e^{-frac{gamma}{m}t}$$
Use your numbers and calculate
answered Mar 7 '17 at 7:45
Kiryl PesotskiKiryl Pesotski
1,734410
$endgroup$
add a comment |
$begingroup$
Let the $x$ axis be along the road, downwards and let the $y$ axis be perpendicular to the road upwards. Projection of gravitational acceleration on the x-axis is $gsin{theta}$ and $-gcos{theta}$ on the y-axis. Friction is along the negative $x$ axis and is $F_{xf}=-gamma{v_{x}}$.
Newton along $x$ is
$$dot{v}_{x}=gsin{theta}-frac{gamma}{m}{v_{x}}$$
the solution is
$${v_{x}}(t)=frac{mg}{gamma}sin{theta}-frac{mc}{gamma}e^{-frac{gamma}{m}t}$$
At $t=0$ the speed was $v_{0x}$, so
$$c=gsin{theta}-frac{gamma}{m}v_{0x}$$
And the solution with initial conditions is
$${v_{x}}(t)=(1-e^{-frac{gamma}{m}t})frac{mg}{gamma}sin{theta}+v_{0x}e^{-frac{gamma}{m}t}$$
Use your numbers and calculate
answered Mar 7 '17 at 7:45
Kiryl PesotskiKiryl Pesotski
1,734410
$endgroup$
Let the $x$ axis be along the road, downwards and let the $y$ axis be perpendicular to the road upwards. Projection of gravitational acceleration on the x-axis is $gsin{theta}$ and $-gcos{theta}$ on the y-axis. Friction is along the negative $x$ axis and is $F_{xf}=-gamma{v_{x}}$.
Newton along $x$ is
$$dot{v}_{x}=gsin{theta}-frac{gamma}{m}{v_{x}}$$
the solution is
$${v_{x}}(t)=frac{mg}{gamma}sin{theta}-frac{mc}{gamma}e^{-frac{gamma}{m}t}$$
At $t=0$ the speed was $v_{0x}$, so
$$c=gsin{theta}-frac{gamma}{m}v_{0x}$$
And the solution with initial conditions is
$${v_{x}}(t)=(1-e^{-frac{gamma}{m}t})frac{mg}{gamma}sin{theta}+v_{0x}e^{-frac{gamma}{m}t}$$
Use your numbers and calculate
answered Mar 7 '17 at 7:45
Kiryl PesotskiKiryl Pesotski
1,734410
answered Mar 7 '17 at 7:45
Kiryl PesotskiKiryl Pesotski
1,734410
answered Mar 7 '17 at 7:45
Kiryl PesotskiKiryl Pesotski
1,734410
answered Mar 7 '17 at 7:45
Kiryl PesotskiKiryl Pesotski
1,734410
1,734410
add a comment |
add a comment |
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$begingroup$
As an expert witness you're probably supposed to base this on actual physics as opposed to naive mathematical models. This therefore becomes more of a physics question. Note for example that the friction for a car is not modelled using coefficient of friction.
$endgroup$
– skyking
Mar 7 '17 at 7:53
$begingroup$
@skyking This is obviously an artificial word problem that involves a fictional and irrelevant expert witness for the story.
$endgroup$
– amd
Mar 7 '17 at 8:39