Two cars leave an intersection at the same time. Car X is travelling East at 50 km/hr and Car Y is travelling South at 60 km/hr. Find the rate at which the cars are separating after 30 minutes.
X and Y move perpedicular to each other, forming a right triangle.
As such, the distance between the cars is sqrt(X^2 + Y^2), where X = +50 km/hr, Y = +60 km/hr. (These are functions.)
We want the RATE at which these cars are seperating, so take the first derivative of the above equation, then plug in .5 into that equation (30 mins = .5 hours).
what will be the final answer of this
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To find the rate at which the cars are separating after 30 minutes, we need to determine the distance between the two cars at that time.
Let's assume that the intersection is the origin (0,0) on a coordinate plane. Car X is traveling East, so its position after 30 minutes can be calculated as (50 km/hr) * (30 min) = 1500 km. Car Y is traveling South, so its position after 30 minutes can be calculated as (60 km/hr) * (30 min) = 1800 km.
Now, we can use the Pythagorean theorem to find the distance between the two cars. The distance (d) is given by the formula:
d = √((x2 - x1)^2 + (y2 - y1)^2)
Where (x1, y1) and (x2, y2) are the positions of Car X and Car Y, respectively. In this case, (x1, y1) is (0, 1500) and (x2, y2) is (1800, 0). Plugging these values into the formula:
d = √((1800 - 0)^2 + (0 - 1500)^2)
= √(1800^2 + (-1500)^2)
= √(3240000 + 2250000)
= √5490000
≈ 2344.6 km
Therefore, the distance between the two cars after 30 minutes is approximately 2344.6 km.
To find the rate at which they are separating, we can consider their velocities as vectors. The velocity vector of Car X is (50 km/hr, 0 km/hr) since it is only traveling East. The velocity vector of Car Y is (0 km/hr, -60 km/hr) since it is only traveling South.
The rate at which the cars are separating can be found by taking the derivative of the distance formula with respect to time (t):
Rate of separation = d'(t)
Using the chain rule, we can write:
Rate of separation = ∂d/∂x * ∂x/∂t + ∂d/∂y * ∂y/∂t
Since x = 50t and y = -60t, we can substitute these values into the formula and evaluate it at t = 0.5 hours (30 minutes) to find the rate of separation after 30 minutes.
Rate of separation = ∂d/∂x * ∂x/∂t + ∂d/∂y * ∂y/∂t
= (1800/√5490000)(50) + (-1500/√5490000)(-60)
= (1800/2344.6)(50) + (-1500/2344.6)(-60)
≈ 60.43 km/hr
Therefore, the rate at which the cars are separating after 30 minutes is approximately 60.43 km/hr.