A police cruiser, approaching a right –angled intersection from the north is chasing a speeding car that has turned the corner and is now moving straight east. When the cruiser is 0.6 mi north of the intersection and the car 0.8 mi to the east, the police determine with radar that the distance between them and the car is increasing at 20 mph. If the cruiser is moving at 60mph at the instant of measurement, what is the speed of the car?
make a sketch
At a time of t hrs,
let the distance the police car is from the intersection be x miles
let the distance the speeding car is from the intersection be y miles
let the distance between them be d miles
d^2 = x^2 + y^2
2d dd/dt = 2x dx/dt + 2y dy/dt
d dd/dt = x dx/dt + y dy/dt
given: x = .6, y = .8 , dx/dt = -60 , dd/dt = 20
( I made dx/dt negative because the value of x is decreasing)
d^2 = .6^2 + .8^2 = 1
d = 1
1(20) = .6(-60) + .8dy/dt
20 + 36 = .8 dy/dt
dy/dt = 56/.8 = 70
The car was speeding at 70 mph
70 mph
To solve this problem, we can use the concept of relative velocity.
Let's assume the speed of the car is 'x' mph.
We are given the following information:
- The distance between the cruiser and the car is increasing at a rate of 20 mph.
- The speed of the cruiser is 60 mph.
Now, let's consider the situation at the instant of measurement:
The speed of the car is 'x' mph, and the speed of the cruiser is 60 mph. The relative velocity between the cruiser and the car is the difference between their speeds, which is (60 - x) mph.
Now, using the concept of relative velocity, we know that the distance between two objects is their relative velocity multiplied by the time.
In this case, the distance between the cruiser and the car is increasing at a rate of 20 mph, which means the relative distance is 20 mph.
Let's denote the time at the instant of measurement as 't' hours.
So, the equation is:
(60 - x) * t = 20
Simplifying this equation:
60t - xt = 20
Now, let's consider the distances traveled by each object:
- The cruiser travels a distance of 0.6 miles north of the intersection.
- The car travels a distance of 0.8 miles to the east.
Using the formula for distance (distance = speed * time), we can set up two equations:
For the cruiser:
60t = 0.6
For the car:
xt = 0.8
Now, let's solve these equations:
From the equation for the cruiser:
60t = 0.6
t = 0.6 / 60
t = 0.01 hours
From the equation for the car:
xt = 0.8
x * 0.01 = 0.8
x = 0.8 / 0.01
x = 80 mph
Therefore, the speed of the car is 80 mph.
To solve this problem, we can use the concepts of relative motion and the Pythagorean theorem.
Let's assume that the speed of the car is represented by 'c' mph.
We know that the police cruiser is moving north at 60 mph, and the distance between them and the car is increasing at 20 mph.
Since the police cruiser is moving north and the car is moving east, the distance between them can be represented as the diagonal of a right-angled triangle. The vertical side of the triangle represents the distance traveled by the police cruiser, and the horizontal side represents the distance traveled by the car.
Using the Pythagorean theorem, we can write the equation:
(vertical distance)^2 + (horizontal distance)^2 = (diagonal distance)^2
(0.6)^2 + (0.8 + 20t)^2 = (60t)^2
Simplifying the equation:
0.36 + (0.64 + 32t + 400t^2) = 3600t^2
0.64 + 32t + 400t^2 = 3600t^2 - 0.36
399.36t^2 - 32t - 0.28 = 0
Now, we can solve this quadratic equation to find the value of 't' in terms of hours. 't' represents the time in hours elapsed since the instant of measurement.
Once we find the value of 't', we can substitute it back into the equation (0.8 + 20t) to find the distance traveled by the car (horizontal distance).
Finally, by dividing the distance traveled by the car by the time elapsed, we can find its speed 'c' in miles per hour.