Question

Smart people please help :D with working out ~
The dockers bridge passes over the coast freeway. The freeway is 20 metres below the bridge and at right angles to it. A car travelling 50 km per hour on the bridge is directly above another car travelling at 100 km per hour on the freeway. Find how fast the cars will be seperating after 1 minute later.

Answer 1

Initial position of car A: (0,0,20) - 20 comes from the height of the bridge
Initial position of car B: (0,0,0)

After 1 min,
Position of car A: (50/60,0,20) - 50/60 is the distance travelled by this car in 1 min
Position of car B: (0,100/60,0) - 100/60 is the distance travelled by this car in 1 min

Now use the distance formula

Answer 2

The answer is 10.87 km/min .. mind if u could do ur working out?

Answer 3

speed is change in distance over change in time. Making a differential equation out of it, \[ v = \frac{\partial r}{\partial t} = \frac{\partial r}{\partial t} = \frac{\partial \sqrt{x^2 + y^2 + z^2}}{\partial t} = \frac{1}{2\sqrt{x^2 + y^2 + z^2}} \frac{\partial \left( x^2 + y^2 + z^2 \right)}{\partial t}\] \[ = \frac{1}{2\sqrt{x^2 + y^2 + z^2}} \left( 2x \frac{\partial x}{\partial t} + 2y \frac{\partial y}{\partial t} + 2z \frac{\partial z}{\partial t} \right) \] \[ = \frac{1}{2\sqrt{0.833^2 + 1.666^2 + 20^2}} \left( 1.666 \times 0.833 + 3.333 \times 1.666 + 40 \times 0 \right) = 0.17 km/hr = 10.87 km/min \]

Answer 4

haha thx i will try that out later after i wake up :P