In triangle ABC , using Pythagorean theorem
BC = sqrt(AB² + AC²)
r = sqrt(y² + x²) eq-1
taking derivative both side relative to "t"
dr/dt = (1/(2 sqrt(y² + x²) ) ) (2 y (dy/dt) + 2 x (dx/dt))
dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) (2 (0.5) (dy/dt) + 2 (0.5) (dx/dt))
dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) ( v₁ + v₂)
15= (1/(2 sqrt(0.5² + 0.5²) ) ) ( - 30 + v₂)
v₂ = 51.2 m/s
The definition of speed allows to find the result for the speed of the car that moves away from the intersection is:
v= 8.79 mph
The velocity is defined by the variation of the position with respect to time.
[tex]v = \frac{dr}{dt}[/tex]
Where v is the instantaneous velocity, r the position and t in time.
It indicates that car 1 is approaching the interception at 30 mph to the north and car 2 is moving away from the intersection towards the east, in the attachment we see a diagram of the positions of the cars
We use the Pythagorean theorem to find the distance
R = [tex]\sqrt{x^2 +y^2}[/tex]
Let's perform the derivative of this expression to find the velocity
[tex]v=\frac{dR }{dt } = \frac{1}{2} \frac{1}{\sqrt{x^2+y^2} } ( 2 x v_x + 2y v_y)\\vx = \frac{dx}{dt}\\vy= \frac{dy}{dt}[/tex]
Indicate that the speed in a straight line is v = 15 mph, speed vy = 30 mph to North direction.
Let's substitute
15 = [tex]\frac{1}{2} \frac{1}{\sqrt{0.5^2+0.5^2} } ( 2 \ 0.5 vx + 2 \ 0.5 \ 30)[/tex]
15 1.4142 = vₓ + 30
vₓ = 121.2 - 30
vₓ = -8.79 mph
The negative sign indicates that the cars moving away
In conclusion with the definition of velocity we can find the velocity of the car that moves away from the intersection is:
8.79 mph
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