"Two cars are traveling along perpendicular roads, car A at 40mph, car B at 60mph. At noon, when car A reaches the intersection, car B is 90miles away, and moving toward it. At 1PM the distance between the cars is changing at what rate?"
So lost! Please help!
The distance d between A and B at noon is 90
At time t thereafter,
d^2 = (90-60t)^2 + (40t)^2
At 1:00, d^2 = 30^2 + 40^2 = 50^2, so d = 50
2d d' = 2(90-60t)*(-60) + 2(40t)*(40)
100 d' = 2(30)(-60) + 2(40)(40) = -3600 + 3200 = -400
d' = -4
so, the cars are moving closer at 4 mph.
FYI, d'(1.038) = 0, so at 1:02 the distance starts to increase.
To find the rate at which the distance between the cars is changing at 1 PM, we need to compute the derivative of the distance function with respect to time. Let's break down the problem step by step.
Let's assume that car A is traveling along the x-axis, and car B is traveling along the y-axis. The position of car A at any time t can be represented as (40t, 0) (since it is moving along the x-axis with a constant speed of 40 mph). Similarly, the position of car B at any time t can be represented as (0, 60t) (since it is moving along the y-axis with a constant speed of 60 mph).
The distance between the two cars at any time t can be calculated using the formula for the distance between two points in a plane:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Distance = sqrt((40t - 0)^2 + (0 - 60t)^2)
Distance = sqrt(1600t^2 + 3600t^2)
Distance = sqrt(5200t^2)
Distance = 20t*sqrt(13)
Now, we can find the rate at which the distance is changing with respect to time by differentiating the distance function with respect to t:
d(Distance)/dt = d(20t*sqrt(13))/dt
= 20*sqrt(13)*dt/dt
= 20*sqrt(13)
Therefore, the rate at which the distance between the cars is changing at 1 PM is 20*sqrt(13) miles per hour.
To find the rate at which the distance between the cars is changing at 1 PM, we can use the concept of relative motion. Let's break down the problem step by step:
Step 1: Determine the distance traveled by car A from noon to 1 PM.
Since car A is traveling at a constant speed of 40 mph, the distance it travels in 1 hour (from noon to 1 PM) is 40 miles.
Step 2: Calculate the distance between the cars at 1 PM.
At noon, the distance between the cars is given as 90 miles. Car A has traveled an additional 40 miles from noon to 1 PM. Therefore, at 1 PM, the distance between the cars will be 90 miles - 40 miles = 50 miles.
Step 3: Find the rate at which the distance between the cars is changing at 1 PM.
To find the rate at which the distance between the cars is changing, we need to calculate the relative velocity between the two cars. Since car B is moving toward car A, the relative velocity can be found by subtracting the speed of car A from the speed of car B.
Relative velocity = Speed of car B - Speed of car A = 60 mph - 40 mph = 20 mph.
Therefore, the distance between the cars is changing at a rate of 20 miles per hour at 1 PM.