i got 5 m/s but it seems to be wrong..can anyone help me out?

Question

2 roads intersects at right angles, one going north-south, the other east-west. An observer stands on the road 60 meters south of the intersection and watches a cyclist traveling east at 10 meters per second. At what rate is the cyclist moving away from the observer 8 seconds after he passes thru the intersection? ( hint: 3-4-5 right angled triangle)

Answer 1

Draw the triangle representing the observer, the intersection and the cyclist. Since it is a right triangle and the two sides are 60m and 10×8 = 80m, we know the hypotenuse must be 100m.

Now, draw a line representing the cyclist's direction of motion and a dotted line extending the line from the observer to the cyclist. We want to project the cyclist's velocity onto this line. That is, go along the dotted line until you find the point where you can draw a line perpendicular to the dotted line that connectes with the end of the line you drew for the cyclist's direction of motion.

What we are doing here is resolving the cyclist's direction of motion into a component directly away from the observer and a component going across. The component directly away from the observer is the one we want.

If you've drawn the diagram correctly, you should have a pair of vertically opposite (and hence equal) angles, call them θ. From the distance triangle you get cos θ = 80/100 = 4/5, and from the direction triangle you get cos θ = x / 10 where x is the speed of the cyclist in the direction away from the observer. From this it follows that x = 8 m/s.

This is actually easier to do with calculus, but given the hint I assume you're not supposed to do it that way. But if you do know calculus, it goes like this:
The distance from the observer at any time t is given by D = √(60^2 + (10t)^2)
so dD/dt = (1/2) (3600 + 100t^2)^(-1/2) (200t)
and at t = 8 this is (1/2) (1/100) (1600) = 8 m/s.

Answer 2

Well I did answer your old question about this, but if its wrong, maybe you should try using a different point to start with. Like 8 seconds and 9 seconds, instead of at intersection and 8 seconds.
Anyways using all the other stuff i told you already the cyclist is 100 meters away from the observer after 8 seconds.
After 9 seconds he is 108.166 meters away.
So the answer might be 8.166 meters/second.
Hope I'm helping

Answer 3

Distance of cyclist from intersection = x
Distance of observer from cyclist = y
y^2 = 60^2 + x^2
2ydy/dx = 2x
dy/dx = x/y = x/(sqrt(60^2+x^2))
dx/dt = 10m/s
dy/dt = dy/dx * dx/dt = 10x/sqrt(60^2 + x^2))
x = 8*10 = 80m
dy/dt = 800/sqrt(60^2+80^2)
= 800/100
= 8m/s

Answer 4

s^2 = x^2 + y^2
2sds/dt = 2xdx/dt + 2ydy/dt
ds/dt = (xdx/dt + ydy/dt)/s
At t = 8 sec
ds/dt = (10*8*10 + 60*0)/√(60^2 + 80^2)
ds/dt = 800/100 = 8 m/s