The resistance of a car problems
$begingroup$
I came across this question:
The resistance of a car whose mass is $750$kg is proportional to its speed. When climbing a slope of $sin^{-1}(1/25)$ at a constant speed of $10$m/s, the engine works at a rate of $80$kW.
When the car is traveling down the same slope with the engine still working at a constant rate of $30$kW, it has speed $v$m/s at time $t$ seconds after starting it's descend.
Show that
$$25frac{dv}{dt}= frac{(10+v)(100-9v)}{v}$$
I know $P = frac{dE}{dt}$. And since the body is moving on a straight line, I assume it possess a kinetic energy thus $frac{mv^2}{2} = pt$. How will I proceed to solve this question?
ordinary-differential-equations
edited Dec 7 '18 at 15:53
amWhy
1
asked Dec 6 '18 at 20:52
AstatineAstatine
86
$endgroup$
add a comment |
$begingroup$
I came across this question:
The resistance of a car whose mass is $750$kg is proportional to its speed. When climbing a slope of $sin^{-1}(1/25)$ at a constant speed of $10$m/s, the engine works at a rate of $80$kW.
When the car is traveling down the same slope with the engine still working at a constant rate of $30$kW, it has speed $v$m/s at time $t$ seconds after starting it's descend.
Show that
$$25frac{dv}{dt}= frac{(10+v)(100-9v)}{v}$$
I know $P = frac{dE}{dt}$. And since the body is moving on a straight line, I assume it possess a kinetic energy thus $frac{mv^2}{2} = pt$. How will I proceed to solve this question?
ordinary-differential-equations
edited Dec 7 '18 at 15:53
amWhy
1
asked Dec 6 '18 at 20:52
AstatineAstatine
86
$endgroup$
$begingroup$
I added some more work to confirm the equation, but there is an inconsistency with the 80 kw power when ascending. Should this be 60kw?
$endgroup$
– Phil H
Dec 7 '18 at 15:10
$begingroup$
Shouldn't this belong to physics.stackexchange.com ?
$endgroup$
– Federico
Dec 7 '18 at 15:17
add a comment |
$begingroup$
I came across this question:
The resistance of a car whose mass is $750$kg is proportional to its speed. When climbing a slope of $sin^{-1}(1/25)$ at a constant speed of $10$m/s, the engine works at a rate of $80$kW.
When the car is traveling down the same slope with the engine still working at a constant rate of $30$kW, it has speed $v$m/s at time $t$ seconds after starting it's descend.
Show that
$$25frac{dv}{dt}= frac{(10+v)(100-9v)}{v}$$
I know $P = frac{dE}{dt}$. And since the body is moving on a straight line, I assume it possess a kinetic energy thus $frac{mv^2}{2} = pt$. How will I proceed to solve this question?
ordinary-differential-equations
edited Dec 7 '18 at 15:53
amWhy
1
asked Dec 6 '18 at 20:52
AstatineAstatine
86
$endgroup$
I came across this question:
The resistance of a car whose mass is $750$kg is proportional to its speed. When climbing a slope of $sin^{-1}(1/25)$ at a constant speed of $10$m/s, the engine works at a rate of $80$kW.
When the car is traveling down the same slope with the engine still working at a constant rate of $30$kW, it has speed $v$m/s at time $t$ seconds after starting it's descend.
Show that
$$25frac{dv}{dt}= frac{(10+v)(100-9v)}{v}$$
I know $P = frac{dE}{dt}$. And since the body is moving on a straight line, I assume it possess a kinetic energy thus $frac{mv^2}{2} = pt$. How will I proceed to solve this question?
ordinary-differential-equations
ordinary-differential-equations
edited Dec 7 '18 at 15:53
amWhy
1
asked Dec 6 '18 at 20:52
AstatineAstatine
86
edited Dec 7 '18 at 15:53
amWhy
1
asked Dec 6 '18 at 20:52
AstatineAstatine
86
edited Dec 7 '18 at 15:53
amWhy
1
edited Dec 7 '18 at 15:53
amWhy
1
edited Dec 7 '18 at 15:53
amWhy
1
1
asked Dec 6 '18 at 20:52
AstatineAstatine
86
asked Dec 6 '18 at 20:52
AstatineAstatine
86
asked Dec 6 '18 at 20:52
AstatineAstatine
86
86
$begingroup$
I added some more work to confirm the equation, but there is an inconsistency with the 80 kw power when ascending. Should this be 60kw?
$endgroup$
– Phil H
Dec 7 '18 at 15:10
$begingroup$
Shouldn't this belong to physics.stackexchange.com ?
$endgroup$
– Federico
Dec 7 '18 at 15:17
add a comment |
$begingroup$
I added some more work to confirm the equation, but there is an inconsistency with the 80 kw power when ascending. Should this be 60kw?
$endgroup$
– Phil H
Dec 7 '18 at 15:10
$begingroup$
Shouldn't this belong to physics.stackexchange.com ?
$endgroup$
– Federico
Dec 7 '18 at 15:17
$begingroup$
I added some more work to confirm the equation, but there is an inconsistency with the 80 kw power when ascending. Should this be 60kw?
$endgroup$
– Phil H
Dec 7 '18 at 15:10
$begingroup$
I added some more work to confirm the equation, but there is an inconsistency with the 80 kw power when ascending. Should this be 60kw?
$endgroup$
– Phil H
Dec 7 '18 at 15:10
$begingroup$
Shouldn't this belong to physics.stackexchange.com ?
$endgroup$
– Federico
Dec 7 '18 at 15:17
$begingroup$
Shouldn't this belong to physics.stackexchange.com ?
$endgroup$
– Federico
Dec 7 '18 at 15:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Disclosure: This is not a rigorous proof but it does show consistency for the given equation of acceleration in switching from ascending to descending. Also I took $g = 10 $ m/s$^2$. It may help in providing insight to a better solution involving energy by taking into account the change in PE by reversing direction.
Taking $frac{dv}{dt} = frac{(10+v)(100-9v)}{25v}$ when the car is descending. In which case at $10 $m/s.....
$frac{dv}{dt} = frac{20(10)}{25(10)} = 0.8 $m/s$^2$
When ascending at $10 $m/s....
$frac{dv}{dt} = 0 = 0.8 - 2(frac{F}{m})$
$0.8 - 2frac{mcdot gcdot sinalpha}{m} = 0$
$0.8 - 2(10cdot 0.04) = 0$
$0.8 - 0.8 = 0$
Hence $frac{dv}{dt} = frac{(10+v)(100-9v)}{25v}$ satisfies the change in acceleration from ascending to descending.
Edit: Investigating this a little more when descending
$frac{dv}{dt} = frac{(10+v)(100-9v)}{25v} = 0$ when.....
$(10+v)(100-9v) = 0$
Hence $v = frac{100}{9}$ m/s
At $v = frac{100}{9}$ m/s the energies must balance....
$30 000 = Rv^2 - mgfrac{v}{25}$
$30 000 = frac{10000}{81}R - 750(10)frac{100}{9cdot25}$
$30 000 = frac{10000}{81}R - 3333.3333$
$33333.3333 = frac{10000}{81}R$
$R = 270$
$$frac{dv}{dt} = frac{F}{m} = frac{frac{P}{v}}{m}$$
$$frac{dv}{dt} = frac{frac{30 000 - 270v^2 + 300v}{v}}{750}$$
$$frac{dv}{dt} = frac{1000 - 9v^2 + 10v}{25v}$$
Hence:
$$frac{dv}{dt} = frac{(10 + v)(100 - 9v)}{25v}$$
This equation also works when ascending at $-10$ m/s giving $frac{dv}{dt} = 0$ but is not consistent with a power of $80$ kw
answered Dec 7 '18 at 0:24
Phil HPhil H
4,1532312
$endgroup$
$begingroup$
Thanks. But can you please elaborate further as to how mgv/25 is gotten, taking from 3000 = Rv^2 - mg v/25. How did you come about the v/25. ? Thanks
$endgroup$
– Astatine
Dec 14 '18 at 1:08
$begingroup$
v/25 is the change in height per second due to the slope. Power 30000 = Rate of change of energy = Force times velocity - Rate of change in PE. Force = Rv (resistance proportional to speed) and Rate of change in PE = mgv/25. This part determines the value of the constant of proportionality R =270 between the resistance force and velocity. So Fr = 270v. Once you have this, then figuring acceleration as a function of velocity is easy.
$endgroup$
– Phil H
Dec 14 '18 at 5:04
$begingroup$
Thanks for the detailed information
$endgroup$
– Astatine
Dec 15 '18 at 10:44
$begingroup$
Hello prof, I have been asked to find the time for the car to reach the speed of 10m/s from rest on it's path down the slope. I try solving it but had an answer of 0.394, please is this correct, if not please elaborate.
$endgroup$
– Astatine
Dec 15 '18 at 13:04
$begingroup$
I used an on-line anti-derivative solver to get a general solution and then I figured the specific solution to be t = (-250ln(|9v - 100|) - 225(|v + 10|))/171 + 9.7624 and for the definite integral between v=0 and v=10 I get t = 2.4543 seconds.
$endgroup$
– Phil H
Dec 15 '18 at 20:35
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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$begingroup$
Disclosure: This is not a rigorous proof but it does show consistency for the given equation of acceleration in switching from ascending to descending. Also I took $g = 10 $ m/s$^2$. It may help in providing insight to a better solution involving energy by taking into account the change in PE by reversing direction.
Taking $frac{dv}{dt} = frac{(10+v)(100-9v)}{25v}$ when the car is descending. In which case at $10 $m/s.....
$frac{dv}{dt} = frac{20(10)}{25(10)} = 0.8 $m/s$^2$
When ascending at $10 $m/s....
$frac{dv}{dt} = 0 = 0.8 - 2(frac{F}{m})$
$0.8 - 2frac{mcdot gcdot sinalpha}{m} = 0$
$0.8 - 2(10cdot 0.04) = 0$
$0.8 - 0.8 = 0$
Hence $frac{dv}{dt} = frac{(10+v)(100-9v)}{25v}$ satisfies the change in acceleration from ascending to descending.
Edit: Investigating this a little more when descending
$frac{dv}{dt} = frac{(10+v)(100-9v)}{25v} = 0$ when.....
$(10+v)(100-9v) = 0$
Hence $v = frac{100}{9}$ m/s
At $v = frac{100}{9}$ m/s the energies must balance....
$30 000 = Rv^2 - mgfrac{v}{25}$
$30 000 = frac{10000}{81}R - 750(10)frac{100}{9cdot25}$
$30 000 = frac{10000}{81}R - 3333.3333$
$33333.3333 = frac{10000}{81}R$
$R = 270$
$$frac{dv}{dt} = frac{F}{m} = frac{frac{P}{v}}{m}$$
$$frac{dv}{dt} = frac{frac{30 000 - 270v^2 + 300v}{v}}{750}$$
$$frac{dv}{dt} = frac{1000 - 9v^2 + 10v}{25v}$$
Hence:
$$frac{dv}{dt} = frac{(10 + v)(100 - 9v)}{25v}$$
This equation also works when ascending at $-10$ m/s giving $frac{dv}{dt} = 0$ but is not consistent with a power of $80$ kw
answered Dec 7 '18 at 0:24
Phil HPhil H
4,1532312
$endgroup$
$begingroup$
Thanks. But can you please elaborate further as to how mgv/25 is gotten, taking from 3000 = Rv^2 - mg v/25. How did you come about the v/25. ? Thanks
$endgroup$
– Astatine
Dec 14 '18 at 1:08
$begingroup$
v/25 is the change in height per second due to the slope. Power 30000 = Rate of change of energy = Force times velocity - Rate of change in PE. Force = Rv (resistance proportional to speed) and Rate of change in PE = mgv/25. This part determines the value of the constant of proportionality R =270 between the resistance force and velocity. So Fr = 270v. Once you have this, then figuring acceleration as a function of velocity is easy.
$endgroup$
– Phil H
Dec 14 '18 at 5:04
$begingroup$
Thanks for the detailed information
$endgroup$
– Astatine
Dec 15 '18 at 10:44
$begingroup$
Hello prof, I have been asked to find the time for the car to reach the speed of 10m/s from rest on it's path down the slope. I try solving it but had an answer of 0.394, please is this correct, if not please elaborate.
$endgroup$
– Astatine
Dec 15 '18 at 13:04
$begingroup$
I used an on-line anti-derivative solver to get a general solution and then I figured the specific solution to be t = (-250ln(|9v - 100|) - 225(|v + 10|))/171 + 9.7624 and for the definite integral between v=0 and v=10 I get t = 2.4543 seconds.
$endgroup$
– Phil H
Dec 15 '18 at 20:35
add a comment |
$begingroup$
Disclosure: This is not a rigorous proof but it does show consistency for the given equation of acceleration in switching from ascending to descending. Also I took $g = 10 $ m/s$^2$. It may help in providing insight to a better solution involving energy by taking into account the change in PE by reversing direction.
Taking $frac{dv}{dt} = frac{(10+v)(100-9v)}{25v}$ when the car is descending. In which case at $10 $m/s.....
$frac{dv}{dt} = frac{20(10)}{25(10)} = 0.8 $m/s$^2$
When ascending at $10 $m/s....
$frac{dv}{dt} = 0 = 0.8 - 2(frac{F}{m})$
$0.8 - 2frac{mcdot gcdot sinalpha}{m} = 0$
$0.8 - 2(10cdot 0.04) = 0$
$0.8 - 0.8 = 0$
Hence $frac{dv}{dt} = frac{(10+v)(100-9v)}{25v}$ satisfies the change in acceleration from ascending to descending.
Edit: Investigating this a little more when descending
$frac{dv}{dt} = frac{(10+v)(100-9v)}{25v} = 0$ when.....
$(10+v)(100-9v) = 0$
Hence $v = frac{100}{9}$ m/s
At $v = frac{100}{9}$ m/s the energies must balance....
$30 000 = Rv^2 - mgfrac{v}{25}$
$30 000 = frac{10000}{81}R - 750(10)frac{100}{9cdot25}$
$30 000 = frac{10000}{81}R - 3333.3333$
$33333.3333 = frac{10000}{81}R$
$R = 270$
$$frac{dv}{dt} = frac{F}{m} = frac{frac{P}{v}}{m}$$
$$frac{dv}{dt} = frac{frac{30 000 - 270v^2 + 300v}{v}}{750}$$
$$frac{dv}{dt} = frac{1000 - 9v^2 + 10v}{25v}$$
Hence:
$$frac{dv}{dt} = frac{(10 + v)(100 - 9v)}{25v}$$
This equation also works when ascending at $-10$ m/s giving $frac{dv}{dt} = 0$ but is not consistent with a power of $80$ kw
answered Dec 7 '18 at 0:24
Phil HPhil H
4,1532312
$endgroup$
$begingroup$
Thanks. But can you please elaborate further as to how mgv/25 is gotten, taking from 3000 = Rv^2 - mg v/25. How did you come about the v/25. ? Thanks
$endgroup$
– Astatine
Dec 14 '18 at 1:08
$begingroup$
v/25 is the change in height per second due to the slope. Power 30000 = Rate of change of energy = Force times velocity - Rate of change in PE. Force = Rv (resistance proportional to speed) and Rate of change in PE = mgv/25. This part determines the value of the constant of proportionality R =270 between the resistance force and velocity. So Fr = 270v. Once you have this, then figuring acceleration as a function of velocity is easy.
$endgroup$
– Phil H
Dec 14 '18 at 5:04
$begingroup$
Thanks for the detailed information
$endgroup$
– Astatine
Dec 15 '18 at 10:44
$begingroup$
Hello prof, I have been asked to find the time for the car to reach the speed of 10m/s from rest on it's path down the slope. I try solving it but had an answer of 0.394, please is this correct, if not please elaborate.
$endgroup$
– Astatine
Dec 15 '18 at 13:04
$begingroup$
I used an on-line anti-derivative solver to get a general solution and then I figured the specific solution to be t = (-250ln(|9v - 100|) - 225(|v + 10|))/171 + 9.7624 and for the definite integral between v=0 and v=10 I get t = 2.4543 seconds.
$endgroup$
– Phil H
Dec 15 '18 at 20:35
add a comment |
$begingroup$
Disclosure: This is not a rigorous proof but it does show consistency for the given equation of acceleration in switching from ascending to descending. Also I took $g = 10 $ m/s$^2$. It may help in providing insight to a better solution involving energy by taking into account the change in PE by reversing direction.
Taking $frac{dv}{dt} = frac{(10+v)(100-9v)}{25v}$ when the car is descending. In which case at $10 $m/s.....
$frac{dv}{dt} = frac{20(10)}{25(10)} = 0.8 $m/s$^2$
When ascending at $10 $m/s....
$frac{dv}{dt} = 0 = 0.8 - 2(frac{F}{m})$
$0.8 - 2frac{mcdot gcdot sinalpha}{m} = 0$
$0.8 - 2(10cdot 0.04) = 0$
$0.8 - 0.8 = 0$
Hence $frac{dv}{dt} = frac{(10+v)(100-9v)}{25v}$ satisfies the change in acceleration from ascending to descending.
Edit: Investigating this a little more when descending
$frac{dv}{dt} = frac{(10+v)(100-9v)}{25v} = 0$ when.....
$(10+v)(100-9v) = 0$
Hence $v = frac{100}{9}$ m/s
At $v = frac{100}{9}$ m/s the energies must balance....
$30 000 = Rv^2 - mgfrac{v}{25}$
$30 000 = frac{10000}{81}R - 750(10)frac{100}{9cdot25}$
$30 000 = frac{10000}{81}R - 3333.3333$
$33333.3333 = frac{10000}{81}R$
$R = 270$
$$frac{dv}{dt} = frac{F}{m} = frac{frac{P}{v}}{m}$$
$$frac{dv}{dt} = frac{frac{30 000 - 270v^2 + 300v}{v}}{750}$$
$$frac{dv}{dt} = frac{1000 - 9v^2 + 10v}{25v}$$
Hence:
$$frac{dv}{dt} = frac{(10 + v)(100 - 9v)}{25v}$$
This equation also works when ascending at $-10$ m/s giving $frac{dv}{dt} = 0$ but is not consistent with a power of $80$ kw
answered Dec 7 '18 at 0:24
Phil HPhil H
4,1532312
$endgroup$
Disclosure: This is not a rigorous proof but it does show consistency for the given equation of acceleration in switching from ascending to descending. Also I took $g = 10 $ m/s$^2$. It may help in providing insight to a better solution involving energy by taking into account the change in PE by reversing direction.
Taking $frac{dv}{dt} = frac{(10+v)(100-9v)}{25v}$ when the car is descending. In which case at $10 $m/s.....
$frac{dv}{dt} = frac{20(10)}{25(10)} = 0.8 $m/s$^2$
When ascending at $10 $m/s....
$frac{dv}{dt} = 0 = 0.8 - 2(frac{F}{m})$
$0.8 - 2frac{mcdot gcdot sinalpha}{m} = 0$
$0.8 - 2(10cdot 0.04) = 0$
$0.8 - 0.8 = 0$
Hence $frac{dv}{dt} = frac{(10+v)(100-9v)}{25v}$ satisfies the change in acceleration from ascending to descending.
Edit: Investigating this a little more when descending
$frac{dv}{dt} = frac{(10+v)(100-9v)}{25v} = 0$ when.....
$(10+v)(100-9v) = 0$
Hence $v = frac{100}{9}$ m/s
At $v = frac{100}{9}$ m/s the energies must balance....
$30 000 = Rv^2 - mgfrac{v}{25}$
$30 000 = frac{10000}{81}R - 750(10)frac{100}{9cdot25}$
$30 000 = frac{10000}{81}R - 3333.3333$
$33333.3333 = frac{10000}{81}R$
$R = 270$
$$frac{dv}{dt} = frac{F}{m} = frac{frac{P}{v}}{m}$$
$$frac{dv}{dt} = frac{frac{30 000 - 270v^2 + 300v}{v}}{750}$$
$$frac{dv}{dt} = frac{1000 - 9v^2 + 10v}{25v}$$
Hence:
$$frac{dv}{dt} = frac{(10 + v)(100 - 9v)}{25v}$$
This equation also works when ascending at $-10$ m/s giving $frac{dv}{dt} = 0$ but is not consistent with a power of $80$ kw
answered Dec 7 '18 at 0:24
Phil HPhil H
4,1532312
edited Dec 7 '18 at 15:08
answered Dec 7 '18 at 0:24
Phil HPhil H
4,1532312
answered Dec 7 '18 at 0:24
Phil HPhil H
4,1532312
answered Dec 7 '18 at 0:24
Phil HPhil H
4,1532312
4,1532312
$begingroup$
Thanks. But can you please elaborate further as to how mgv/25 is gotten, taking from 3000 = Rv^2 - mg v/25. How did you come about the v/25. ? Thanks
$endgroup$
– Astatine
Dec 14 '18 at 1:08
$begingroup$
v/25 is the change in height per second due to the slope. Power 30000 = Rate of change of energy = Force times velocity - Rate of change in PE. Force = Rv (resistance proportional to speed) and Rate of change in PE = mgv/25. This part determines the value of the constant of proportionality R =270 between the resistance force and velocity. So Fr = 270v. Once you have this, then figuring acceleration as a function of velocity is easy.
$endgroup$
– Phil H
Dec 14 '18 at 5:04
$begingroup$
Thanks for the detailed information
$endgroup$
– Astatine
Dec 15 '18 at 10:44
$begingroup$
Hello prof, I have been asked to find the time for the car to reach the speed of 10m/s from rest on it's path down the slope. I try solving it but had an answer of 0.394, please is this correct, if not please elaborate.
$endgroup$
– Astatine
Dec 15 '18 at 13:04
$begingroup$
I used an on-line anti-derivative solver to get a general solution and then I figured the specific solution to be t = (-250ln(|9v - 100|) - 225(|v + 10|))/171 + 9.7624 and for the definite integral between v=0 and v=10 I get t = 2.4543 seconds.
$endgroup$
– Phil H
Dec 15 '18 at 20:35
add a comment |
$begingroup$
Thanks. But can you please elaborate further as to how mgv/25 is gotten, taking from 3000 = Rv^2 - mg v/25. How did you come about the v/25. ? Thanks
$endgroup$
– Astatine
Dec 14 '18 at 1:08
$begingroup$
v/25 is the change in height per second due to the slope. Power 30000 = Rate of change of energy = Force times velocity - Rate of change in PE. Force = Rv (resistance proportional to speed) and Rate of change in PE = mgv/25. This part determines the value of the constant of proportionality R =270 between the resistance force and velocity. So Fr = 270v. Once you have this, then figuring acceleration as a function of velocity is easy.
$endgroup$
– Phil H
Dec 14 '18 at 5:04
$begingroup$
Thanks for the detailed information
$endgroup$
– Astatine
Dec 15 '18 at 10:44
$begingroup$
Hello prof, I have been asked to find the time for the car to reach the speed of 10m/s from rest on it's path down the slope. I try solving it but had an answer of 0.394, please is this correct, if not please elaborate.
$endgroup$
– Astatine
Dec 15 '18 at 13:04
$begingroup$
I used an on-line anti-derivative solver to get a general solution and then I figured the specific solution to be t = (-250ln(|9v - 100|) - 225(|v + 10|))/171 + 9.7624 and for the definite integral between v=0 and v=10 I get t = 2.4543 seconds.
$endgroup$
– Phil H
Dec 15 '18 at 20:35
$begingroup$
Thanks. But can you please elaborate further as to how mgv/25 is gotten, taking from 3000 = Rv^2 - mg v/25. How did you come about the v/25. ? Thanks
$endgroup$
– Astatine
Dec 14 '18 at 1:08
$begingroup$
Thanks. But can you please elaborate further as to how mgv/25 is gotten, taking from 3000 = Rv^2 - mg v/25. How did you come about the v/25. ? Thanks
$endgroup$
– Astatine
Dec 14 '18 at 1:08
$begingroup$
v/25 is the change in height per second due to the slope. Power 30000 = Rate of change of energy = Force times velocity - Rate of change in PE. Force = Rv (resistance proportional to speed) and Rate of change in PE = mgv/25. This part determines the value of the constant of proportionality R =270 between the resistance force and velocity. So Fr = 270v. Once you have this, then figuring acceleration as a function of velocity is easy.
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– Phil H
Dec 14 '18 at 5:04
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v/25 is the change in height per second due to the slope. Power 30000 = Rate of change of energy = Force times velocity - Rate of change in PE. Force = Rv (resistance proportional to speed) and Rate of change in PE = mgv/25. This part determines the value of the constant of proportionality R =270 between the resistance force and velocity. So Fr = 270v. Once you have this, then figuring acceleration as a function of velocity is easy.
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– Phil H
Dec 14 '18 at 5:04
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Thanks for the detailed information
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– Astatine
Dec 15 '18 at 10:44
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Thanks for the detailed information
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– Astatine
Dec 15 '18 at 10:44
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Hello prof, I have been asked to find the time for the car to reach the speed of 10m/s from rest on it's path down the slope. I try solving it but had an answer of 0.394, please is this correct, if not please elaborate.
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– Astatine
Dec 15 '18 at 13:04
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Hello prof, I have been asked to find the time for the car to reach the speed of 10m/s from rest on it's path down the slope. I try solving it but had an answer of 0.394, please is this correct, if not please elaborate.
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– Astatine
Dec 15 '18 at 13:04
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I used an on-line anti-derivative solver to get a general solution and then I figured the specific solution to be t = (-250ln(|9v - 100|) - 225(|v + 10|))/171 + 9.7624 and for the definite integral between v=0 and v=10 I get t = 2.4543 seconds.
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– Phil H
Dec 15 '18 at 20:35
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I used an on-line anti-derivative solver to get a general solution and then I figured the specific solution to be t = (-250ln(|9v - 100|) - 225(|v + 10|))/171 + 9.7624 and for the definite integral between v=0 and v=10 I get t = 2.4543 seconds.
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– Phil H
Dec 15 '18 at 20:35
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I added some more work to confirm the equation, but there is an inconsistency with the 80 kw power when ascending. Should this be 60kw?
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– Phil H
Dec 7 '18 at 15:10
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Shouldn't this belong to physics.stackexchange.com ?
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– Federico
Dec 7 '18 at 15:17