I got no idea

[Studyinghard]

A police car is driving northward chasing a vehicle which has just taken a right turn at an
intersection and is travelling due east. When the police car is 600m south of the
intersection, and the car is 800m east of the intersection, an aerial speed radar indicates
that the distance between the two cars is increasing by 20 km/h. If the police car is
travelling at 60km/h when the radar measurement is taken, how fast is the pursued
vehicle going?

[xZero]

MTH1020?

I'll give you some hints, express the distance from the intersection to the cop as a function of time, same for the other car, use them to find the distance between them using pythagoras theorem and diff it which will give you dr/dt. If you're still struggling I'll type up the solution

[brightsky]

Answer should be 70 I think...

So dD/dt = 20
and since one's travelling northwards and one's travelling eastwards, we find the distance through pythag:
D = ((0.6 - 60t)^2 + (0.8 + xt)^2)^(1/2) where x is the speed of the vehicle being chased.
so after some fiddling we get:
dD/dt = 1/2(t^2x^2 + 3600t^2 + 1.6t*x - 72t + 1)^(-1/2) (2x^2t + 7200t + 1.6x - 72) using the chain rule.
but since the radar measurement was taken when t = 0, then we let t = 0.
dD/dt = (1/2(1.6x - 72))/(1)^(1/2) = 1/2(1.6x - 72) = 20
Solve that and we have our answer:
1/2(1.6x - 72) = 20
1.6x - 72 = 40
x = 70

[xZero]

Boo you spoiled it and yeh the answer is 70km/hr

[Studyinghard]

How the hell did you diff that equation i keep getting a different answer

[xZero]

Chain rule or implicite diff.