Two cars are travelling east along a straight highway at the same speed. At an intersection, the highway branches and splits into two straight roads, one heading approximately north-east and one heading approximately south-east. The angle betwen the two roads is 60 degrees. The two cars leave the intersection at the same time, one travelling north-east at 100 km/h and the other travelling south-east at 80 km/h. How fast is the distance between the two cars changing 30 minutes after they leave the intersection.
using the law of cosines, the distance z at time t hiours is
z^2 = (100t)^2 + (80t)^2 - 2(100t)(80t)cos60° = 8400t^2
z = 10√21 t
dz/dt is a constant 10√21 km/hr
so how would I find how fast the distance is changing 30 min after they leave the intersection?
and I got z = 10√84t
To find the rate at which the distance between the two cars is changing, we need to differentiate the distance formula with respect to time and then plug in the given values.
Let's consider the following diagram:
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| C
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| /
| /
| /
| /
| /
| / T
| /
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A | / D
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B
Here, A and B are the positions of the two cars, and C and D are the future positions of the two cars after a time t.
Let's define the following variables:
- x: the distance AB (distance between the two cars)
- y: the distance AC (distance travelled by car 1)
- z: the distance AD (distance travelled by car 2)
We are given that y = 100t (since car 1 is traveling at 100 km/h) and z = 80t (since car 2 is traveling at 80 km/h).
Using trigonometry, we can find that x = √(y^2 + z^2 - 2yzcos(60°)), as it is a triangle with sides y, z, and the angle between them being 60°.
Now, let's differentiate x with respect to t using the chain rule:
dx/dt = (d/dt)√(y^2 + z^2 - 2yzcos(60°))
To simplify the expression, let's differentiate inside the square root:
dx/dt = (1/2)(y^2 + z^2 - 2yzcos(60°))^(-1/2) * (2yy' + 2zz' - 2(cos(60°))(yz'))
Substituting the given values:
dx/dt = (1/2)(100^2 + 80^2 - 2(100)(80)cos(60°))^(-1/2) * (2(100)(100) + 2(80)(80) - 2(cos(60°))(80)(100))
Simplifying further:
dx/dt = (1/2)(10000 + 6400 - 16000*cos(60°))^(-1/2) * (20000 + 12800 - 9600)
dx/dt = (sqrt(25600))^(-1/2) * (23200)
Since it is given that t = 30 minutes (which is equal to 0.5 hours), we can evaluate dx/dt at t = 0.5:
dx/dt = (1/sqrt(25600)) * (23200) ≈ 1.4 km/h
Therefore, the distance between the two cars is changing at a rate of approximately 1.4 km/h after 30 minutes.