1D Kinematics - Integration of the Equations of Motion

[bearjew11]

1. The distance from two airports is 1286 km by air. Plane A leaves the first airport at 10:00a heading north toward the second airport, another plane leaves from the second airport at 11:00a heading south towards the destination plane A originally departed from. Plane A travels at 720km/h, and plane B, slowed by a headwind, travels at 640km/h. Where do the planes meet? At what time?

Given:
Δy= 1286km
V_p1 = 720km/h
V_p2 = 640km/h
Δt = ?
??



2. Not quite sure yet.



3. I tried to draw a graph but it, unfortunately, got me no where.

 

[Chestermiller]

If t is the clock time, how far north has the first plane traveled by time t?
How far south has the second plane traveled by time t?

When the two planes meet, how are their distances from their respective starting points related to the total distance 1286 km?

 

[bearjew11]

If t is the clock time, how far north has the first plane traveled by time t?
How far south has the second plane traveled by time t?

When the two planes meet, how are their distances from their respective starting points related to the total distance 1286 km?

Well, if plane A left an hour before plane B, then the distance is no longer 1286km, (assuming that the position of plane A is 0). The two planes have to, instead, cover 566km to meet, 1286 - 720* 1 = 566km

After 1 hour, or 3600s, plane A has traveled north 720,000m and plane B is beginning to cover distance.

My attempt at a solution:
Knowns and unknowns:
For plane A:
V_A = 720km/h = 200m/s - velocity
x_A = 0km - position
V_B = 640km/h ~~ 178m/s
x_B = 1268km - (720km*1h) = 566km = 566000m
x_m = ? - position where planes meet

Equations used:
t = x_B/[V_A + V_B] = 566000/(200+178) ~~ 1,497.35s - when they meet
x_m = V_A*(t) = 200*(1497.35) = 299,470m - where they meet

 

[Chestermiller]

Nice job.

I think they were asking for the clock time that they meet. 1495.35 sec ~ 25 minutes, so they meet at ~11:25.

Chet

 

[Nickie]

I would approach this problem by using the principles of 1D kinematics and the equations of motion. Firstly, I would identify the known values and variables in the problem. The distance between the two airports (Δy) is given as 1286 km, the initial velocity of plane A (Vp1) is 720 km/h, and the initial velocity of plane B (Vp2) is 640 km/h. The unknown variables are the time (Δt) and the position where the planes meet (Δx).

Next, I would use the equation Δx = Vp1Δt + 1/2aΔt^2 to find the position where the planes meet. Since both planes are moving in opposite directions, the total displacement will be the sum of their individual displacements, which can be represented as Δx = Vp1Δt + Vp2Δt. Substituting the known values, we get 1286 km = (720 km/h + 640 km/h)Δt. Solving for Δt, we get Δt = 1.5 hours.

Therefore, the planes will meet after 1.5 hours of flight time. To find the position where they meet, we can substitute this value of Δt into the equation for displacement, which gives us Δx = (720 km/h)(1.5 hours) + (640 km/h)(1.5 hours) = 1920 km. This means that the planes will meet at a position 1920 km from the first airport.

In conclusion, the planes will meet after 1.5 hours of flight time at a position 1920 km from the first airport. This can be verified by drawing a graph and plotting the positions of the two planes at different times. This problem demonstrates the application of 1D kinematics and the integration of equations of motion in solving real-world problems.