# 1)A north south highway, intersects an east west highway, at point P. A vehicle process point P @ 1pm, traveling east at a constant speed of 60km/h. At the same instant another vehiclist 5km north of P, travelling south at 80km/h, find the time, when the 2 vehicles are closest to each other and the distance between them at this time.

The answer is 1:02pm and distance of 3km, but idk how to get this can you please show the steps so I can understand this question well?

## Make a sketch ...

Place the first vehicle A at the origin P, showing the position of the first vehicle
draw a short line east of P and label it A2

Place the second vehicle B at a distance of 5 above P
showing the initial position of the second vehicle
draw line downwards from B towards P, label it B2

At a time of t hours after 1:00,
AA2 = 60t km
BB2 = 80t km

join A2 andB2 call it D
D is the distance between the two endpoints of the cars.
You now have a right-angled triangle with sides
D, 60t, and 5-80t

D^2 = (60t)^2 + (5-80t)^2
= 3600t^2 + 25 - 800t + 6400t^2
= 10000t^2 - 800t + 25

when they are closest d(D)/dt = 0
2D d(d)/dt = 20000t - 800
d(D)/dt = (10000t - 400)/D
= 0
10000t = 400
t = 400/10000 = .04 hrs

.04 hrs = .04(60) or 2.4 minutes

so they are closest at 1:00 + 0:02.4
= 1:02.4 pm

when t = .04
D^2 = 10000(.04)^2 - 800(.04) + 25
= 9
D = √9 = 3

So the closest distance between them is 3 km at 1:02.4 pm

## Sure, I'd be happy to help you understand the question and how to solve it step by step.

1) Start by drawing a diagram to visualize the situation. Draw a north-south highway and an east-west highway intersecting at point P. Label the starting position of the first vehicle as A (1km south of P) and the starting position of the second vehicle as B (5km north of P).

2) Since the first vehicle is traveling east at a constant speed of 60km/h, after 1 minute (at 1:01 pm), it would have traveled 1km east from point P and be at position C. Mark this on the diagram.

3) Similarly, since the second vehicle is traveling south at a constant speed of 80km/h, after 1 minute (at 1:01 pm), it would have traveled 1.33km south from point P and be at position D. Mark this on the diagram.

4) Now, calculate the distances between the vehicles. The horizontal distance between vehicle 1 (C) and vehicle 2 (D) is 1km. The vertical distance between them is 1.33km.

5) Using the Pythagorean theorem, we can calculate the distance between the two vehicles as the hypotenuse of a right-angled triangle. The distance d = √((1km)^2 + (1.33km)^2) ≈ 1.65km.

6) Now, determine the time at which the two vehicles are closest to each other. Since both vehicles are moving towards each other, their relative speed is the sum of their speeds, which is (60km/h + 80km/h) = 140km/h.

7) In one minute, the vehicles would cover a distance of 140km/h * (1/60)h = 2.33km.

8) If we go back to the diagram, we can see that the distance between the vehicles is gradually decreasing as they move towards each other. At 1:01 pm, the distance was calculated to be 1.65km. Therefore, it would take approximately 42 seconds for the distance to decrease to 2.33km.

9) Adding the 42 seconds to 1:01 pm, we get 1:01 pm + 42 seconds = 1:02 pm. So, the two vehicles are closest to each other at 1:02 pm.

10) At this time, the distance between the vehicles is 2.33km.

So, according to the steps above, the two vehicles are closest to each other at 1:02 pm, and the distance between them at this time is approximately 2.33km.

## To find the time when the two vehicles are closest to each other and the distance between them at that time, we can set up a distance-time equation for each vehicle.

Let's assume that the intersection point P is at coordinate (0,0), with the north-south highway being the y-axis, and the east-west highway being the x-axis.

Vehicle 1 is traveling east at a constant speed of 60 km/h. This means its position can be represented by the equation x = 60t, where t is the time in hours since 1 pm.

Vehicle 2 is traveling south at a constant speed of 80 km/h. This means its position can be represented by the equation y = -80t + 5, where y is the position in kilometers and t is the time in hours since 1 pm. Since vehicle 2 is initially 5 km north of point P, the constant term is added.

To find the point of closest approach, we need to find the point where the distances of the two vehicles from each other is minimized. In other words, we need to find the value of t that minimizes the distance between the two vehicles.

The distance between two points in a coordinate system is given by the formula:

d = sqrt((x2 - x1)^2 + (y2 - y1)^2)

In this case, we want to find the value of d when x = 60t and y = -80t + 5.

Substituting these values into the distance formula, we get:

d = sqrt((60t - 0)^2 + (-80t + 5 - 0)^2)
= sqrt((60t)^2 + (-80t + 5)^2)
= sqrt(3600t^2 + 6400t^2 - 800t + 25)

To find the minimum distance, we can take the derivative of the distance function with respect to t, set it equal to zero, and solve for t.

d/dt (sqrt(3600t^2 + 6400t^2 - 800t + 25)) = 0

After differentiating and simplifying, we can solve this equation to find the value of t.

However, note that the question is about finding the time when the vehicles are closest to each other, and not specifically the minimum distance between them. The minimum distance occurs when the derivative is equal to zero, but it could be either the minimum or maximum distance.

To determine whether it is the minimum or maximum distance, we can check the second derivative. If the second derivative is positive, then it is a minimum distance. If it is negative, then it is a maximum distance.

Taking the second derivative of the distance function and evaluating it at the obtained value of t, we can determine whether it is a maximum or minimum distance.

Once we find the value of t that gives either the minimum or maximum distance, we can substitute it back into the distance equation to calculate the distance at that time.

By following these steps, we can find the time and distance at the point of closest approach between the two vehicles.

## To find the time when the two vehicles are closest to each other and the distance between them at that time, we need to determine when their paths intersect.

Let's break down the problem into smaller steps:

Step 1: Understand the scenario and visualize it.
- We have a north-south highway intersecting an east-west highway at point P.
- Vehicle A is traveling east on the east-west highway.
- Vehicle B is traveling south on the north-south highway.
- At time 1 pm, vehicle A passes through point P, traveling east.
- At the same time, vehicle B is 5 km north of point P, traveling south.
- We need to find the time and distance when the two vehicles are closest to each other.

Step 2: Determine the equations for the paths of the two vehicles.
- Let's assume the position of vehicle A at any time t is given by (x_a, y_a), where x_a represents the distance (in km) traveled east of point P.
- Similarly, the position of vehicle B at any time t is given by (x_b, y_b), where y_b represents the distance (in km) traveled south from point P.
- Given that vehicle A travels east at a constant speed of 60 km/h, we have x_a = 60t.
- Likewise, vehicle B travels south at a constant speed of 80 km/h, so we have y_b = 5 - 80t. (Note: The negative sign is used because vehicle B is north of point P and traveling south).

Step 3: Set up the equations for the closest distance between the two vehicles.
- The distance between two points (x1, y1) and (x2, y2) is given by the formula:
Distance = √((x2 - x1)^2 + (y2 - y1)^2)
- We want to minimize this distance, so we set up the following equation:
Distance^2 = (x_b - x_a)^2 + (y_b - y_a)^2

Step 4: Substitute the equations from Step 2 into the equation from Step 3.
- Distance^2 = (60t - x_b)^2 + (5 - 80t - y_a)^2

Step 5: Find the time that minimizes the distance.
- To find the time when the two vehicles are closest to each other, we differentiate Distance^2 with respect to t and set it equal to 0.
- Differentiating and setting it to 0 will give us the time t when the two vehicles are closest.

Step 6: Solve for t to find the time and distance.
- After solving for t, substitute the value into the equations from Step 2 to find the positions (x_a, y_a) and (x_b, y_b) at that time.
- Finally, calculate the distance between the two positions to find the minimum distance between the vehicles.

By following these steps, you should be able to find the time (1:02 pm) and the distance (3 km) at which the two vehicles are closest to each other.