dynamics help! (1 Viewer)

VJ30 · Aug 19, 2012

a car is travelling along a horizontal road with a velocity of 30 m/s. the engine is stopped at a marked point O along the road and then the car coasts to a stop at a point P. Frictional resistance is 1/10 of the weight of the car and air resistance is v/20 per unit mass, where v is the speed of the car. Find the distance OP and the time to travel this distance. (Take g=10)

thanks!

VJ30 · Aug 19, 2012

VJ30 · Aug 19, 2012

nightweaver066 · Aug 19, 2012

VJ30 said:
a car is travelling along a horizontal road with a velocity of 30 m/s. the engine is stopped at a marked point O along the road and then the car coasts to a stop at a point P. Frictional resistance is 1/10 of the weight of the car and air resistance is v/20 per unit mass, where v is the speed of the car. Find the distance OP and the time to travel this distance. (Take g=10)

thanks!

By Newton's 2nd law of motion,

$ma = -(\frac{1}{10}mg + \frac{mv}{20})$

$= -\frac{m}{20}(20 + v)$

$a = -\frac{1}{20}(20 + v)$

$v\frac{dv}{dx} = -\frac{1}{20}(20 + v)$

$\int ^0 _{30} \frac{v}{20 + v}dv = \int ^d _0 -\frac{1}{20}dx$

$\frac{1}{20}d = \int ^{30} _0 1 - \frac{20}{20 + v}dv$

$d = 20[v - 20ln|20 + v|]^{30}_0$

$= 20(30 - 20ln50 - (-20ln20))$

$= 20(30 + 20ln\frac{2}{5})m$

$\therefore OP = 600 + 400ln\frac{2}{5}m$

Going back to $a = -\frac{1}{20}(20 + v)$

$\frac{dv}{dt} = -\frac{1}{20}(20 + v)$

$\int_{30} ^0 \frac{1}{20 + v}dv = \int _0 ^T -\frac{1}{20}dt$

$\frac{1}{20}T = [ln|20 + v|]^{30}_0$

$T = 20(ln50 - ln20)$

$\therefore T = 20ln\frac{5}{2} s$

VJ30 · Aug 19, 2012

nightweaver066 said:
By Newton's 2nd law of motion,

$ma = -(\frac{1}{10}mg + \frac{mv}{20})$

$= -\frac{m}{20}(20 + v)$

$a = -\frac{1}{20}(20 + v)$

$v\frac{dv}{dx} = -\frac{1}{20}(20 + v)$

$\int ^0 _{30} \frac{v}{20 + v}dv = \int ^d _0 -\frac{1}{20}dx$

$\frac{1}{20}d = \int ^{30} _0 1 - \frac{20}{20 + v}dv$

$d = [v - 20ln|20 + v|]^_{30}_0$

$= (30 - 20ln50 - (-20ln20))$

$= (30 + ln\frac{2}{5})m$

$\therefore OP = (30 + ln\frac{2}{5})m$

Going back to $a = -\frac{1}{20}(20 + v)$

$\frac{dv}{dt} = -\frac{1}{20}(20 + v)$

$\int_{30} ^0 \frac{1}{20 + v}dv = \int _0 ^T -\frac{1}{20}dt$

$\frac{1}{20}T = [ln|20 + v|]^{30}_0$

$T = 20(ln50 - ln20)$

$\therefore T = 20ln\frac{5}{2} s$

lol when i posted this question, i was 100% sure u will be the first guy to answer it..thanks!

edit- i though air resistance was a force itself, not acceleration

nightweaver066 · Aug 19, 2012

VJ30 said:
lol when i posted this question, i was 100% sure u will be the first guy to answer it..thanks!

edit- i though air resistance was a force itself, not acceleration

Haha just helping around as everyone else is at the BoS meat

It is, that's why i subtracted it when considering the forces, using $\Sigma F = ma$

D94 · Aug 19, 2012

VJ30 said:
lol when i posted this question, i was 100% sure u will be the first guy to answer it..thanks!

edit- i though air resistance was a force itself, not acceleration

In the question, it says: "air resistance is v/20 per unit mass", so it's a force. You need to multiply that by the mass of the car to determine the force of the whole body, so you get mv/20.

VJ30 · Aug 19, 2012

nightweaver066 said:
By Newton's 2nd law of motion,

$ma = -(\frac{1}{10}mg + \frac{mv}{20})$

$= -\frac{m}{20}(20 + v)$

$a = -\frac{1}{20}(20 + v)$

$v\frac{dv}{dx} = -\frac{1}{20}(20 + v)$

$\int ^0 _{30} \frac{v}{20 + v}dv = \int ^d _0 -\frac{1}{20}dx$

$\frac{1}{20}d = \int ^{30} _0 1 - \frac{20}{20 + v}dv$

$d = [v - 20ln|20 + v|]^_{30}_0$

$= (30 - 20ln50 - (-20ln20))$

$= (30 + ln\frac{2}{5})m$

$\therefore OP = (30 + ln\frac{2}{5})m$

Going back to $a = -\frac{1}{20}(20 + v)$

$\frac{dv}{dt} = -\frac{1}{20}(20 + v)$

$\int_{30} ^0 \frac{1}{20 + v}dv = \int _0 ^T -\frac{1}{20}dt$

$\frac{1}{20}T = [ln|20 + v|]^{30}_0$

$T = 20(ln50 - ln20)$

$\therefore T = 20ln\frac{5}{2} s$

again why did u multiply 30+ln2/5 by m, where did it come from??

nightweaver066 · Aug 19, 2012

VJ30 said:
again why did u multiply 30+ln2/5 by m, where did it come from??

Sorry, i meant it to mean metres

VJ30 · Aug 19, 2012

for OP i got 40ln2/5 + 600 ??

nightweaver066 · Aug 19, 2012

VJ30 said:
for OP i got 40ln2/5 + 600 ??

I had a typo before which may have confused you and i've corrected it.

It is 30 + 20ln(2/5)

D94 · Aug 19, 2012

OP should = 400ln(2/5)+600.

D94 · Aug 19, 2012

nightweaver066 said:
I had a typo before which may have confused you and i've corrected it.

It is 30 + 20ln(2/5)

You haven't multiplied by 20. You have 1/20 d on one side, you haven't multiplied that across to find d by itself.

nightweaver066 · Aug 19, 2012

D94 said:
You haven't multiplied by 20. You have 1/20 d on one side, you haven't multiplied that across to find d by itself.

ahh okay thanks, i'll update

VJ30 · Aug 19, 2012

D94 said:
OP should = 400ln(2/5)+600.

i realised my mistake...i got that! thanks

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