# Velocity of separation: How fast is the distance between two cars is changing?

#### WMDhamnekar

Hello,

A Land cruiser drives east from point A at 40 kph. Another car Ford Expedition, starting from B at the same time , drives northwest $30^\circ$ towards A at 60 kph. B is 40 km from A.

How fast in kph is the distance between two cars changing after 47 minutes?

Solution:-

My attempt: I want to use here sine law and cosine law. The computations are somewhat complicated and requires thorough thinking. I am trying to answer this question but it will take some time.

Meanwhile, if any member knowing the correct answer to this question , may reply with correct answer.

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#### WMDhamnekar

My answer to this question is 87.3184 kph . In other words, after 47 minutes, the distance between the two cars are changing at 61.3333 kph.

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#### Cervesa

from your problem statement, I am interpreting that point B is located 40 km, 30 degrees south of east from point A

if that is correct, setting point A at the origin places point B at position $(20\sqrt{3},-20)$, units in km.

position as a function of time in hours for the land cruiser,
$x=40t$ and $y=0$

for the Ford,
$x=20\sqrt{3} -30\sqrt{3} \cdot t$ and $y=30t-20$

let $z$ be the distance at any time t (in hours) between the two vehicles

$z^2 = (40t-20\sqrt{3}+30\sqrt{3} \cdot t)^2 +(20-30t)^2$

$\dfrac{dz}{dt} = \dfrac{[(40+30\sqrt{3})t-20\sqrt{3}] \cdot (40+30\sqrt{3}) -30(20-30t)}{z}$

barring any error on my part, you can evaluate the above expression at $t=\dfrac{47}{60}$

I get approx 94.4 km/hr

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#### WMDhamnekar

Hello,
I want to change my answer to this question to 77.87289 kph after 47 minutes. It is computed as follows:- $\cos{50.127^\circ}*60$kph(speed of Ford Expedition) +$\cos{9.873^\circ}*40$ kph(speed of Land Cruiser)=77.87289.

#### WMDhamnekar

Hello,
After 47 minutes, Ford Expedition will be at point D and Land Cruiser will be at point C, making an angle of $50.127^\circ$ and $9.873^\circ$ to the line CD respectively.So the velocity of separation is as per compuations in #4.

#### Debsta

MHF Helper
I agree with 77.87km/h as the answer. I, also, am interpreting that point B as being located 40 km, 30 degrees south of east from point A.
I used the cosine rule to get a formula for D (distance between the two cars) and then calculus when 47/60.
(WMD, I don't really understand your diagram, I have 30 where you have 60 but didn't find any distances like you did.)

#### WMDhamnekar

Hello,
After 47 minutes, Ford Expedition will be at point D and Land Cruiser will be at point C, making an angle of 50.127∘50.127∘ and 9.873∘9.873∘ to the line CD respectively.So the velocity of separation is as per compuations in #4.
I agree with 77.87km/h as the answer. I, also, am interpreting that point B as being located 40 km, 30 degrees south of east from point A.
I used the cosine rule to get a formula for D (distance between the two cars) and then calculus when 47/60.
(WMD, I don't really understand your diagram, I have 30 where you have 60 but didn't find any distances like you did.)
Hello,

Point B is located $60^\circ$ south east 40km away from the point A. Should i write that from point B, another car Ford Expedition drives $North30^\circ west$ towards A?

But How did you use cosine rule to find out point D? How did you use then calculus? Would you answer this question using your method?

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#### Debsta

MHF Helper
60 degrees south east is so confusing. "South-east" is S 45 E.
Do you mean it is 60 south from east or 60 east from south? In correct notation, is it S 60 E? (This notation always starts with N or S and then swings either E or W.)

#### WMDhamnekar

Hello,
Sorry, I made mistake in notation.

But are you presenting your method of finding answer to this question? Because another member posted wrong answer 94.4 km per hour using calculus.

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#### Cervesa

Hello,
Sorry, I made mistake in notation.

But are you presenting your method of finding answer to this question? Because another member posted wrong answer 94.4 km per hour using calculus.
I stated my assumption of the initial position of point B in my original statement. Since your update, it is clear that I misinterpreted your direction notation which should have been 30 degrees west of north, not 30 degrees northwest.

updating ...

point B is located 40 km, 60 degrees south of east from point A

if that is correct, setting point A at the origin places point B at position $(20, -20\sqrt{3})$, units in km.

position as a function of time in hours for the land cruiser,
$x=40t$ and $y=0$

for the Ford,
$x=20 -30t$ and $y=30\sqrt{3} t - 20\sqrt{3}$

let $z$ be the distance at any time t (in hours) between the two vehicles

$z^2 = (70t-20)^2 +(20\sqrt{3}-30\sqrt{3}t)^2$

$\dfrac{dz}{dt} = \dfrac{(70(70t-20) - 30\sqrt{3}(20\sqrt{3}-30\sqrt{3}t)}{z}$

at $t=\dfrac{47}{60}$, $\dfrac{dz}{dt} \approx 77.87$ km/hr

WMDhamnekar