moving beltway at an airport has a speed v1 and a S length L. A woman stands on the beltway as it moves from one end to the other, while a man in a hurry to reach his flight walks on the beltway with a speed of v2 relative to the moving beltway. (a) What time inter- val is required for the woman to travel the distance L? (b) What time interval is required for the man to travel this distance? (c) A second beltway is located next to the first one. It is identical to the first one but moves in the opposite direction at speed v1. Just as the man steps onto the beginning of the beltway and begins to walk at speed v2 relative to his beltway, a child steps on the other end of the adjacent beltway. The child stands at rest relative to this second beltway. How long after
stepping on the beltway does the man pass the child
For part a its simple : L/v1
For part b the mans velocity will be v1+v2/L
The thing that is confusing me is part C this is what I get:
Man: Xf = Xi + Vxt + 1/2at^2
L = (V2+V1)t
Child : -L = -Vit
(V2+v2)t= Vit
???????? So im confused here
, how do I solve for t?
To solve part (c), you will use the equations of motion for the man and the child on the opposite beltway.
For the man:
Let's assume the distance between the two ends of the adjacent beltways is also L (similar to the first beltway). When the man steps onto the beltway, his initial position (Xi) is 0, and his final position (Xf) is L.
Using the equation of motion:
Xf = Xi + Vxt + 1/2at^2
L = 0 + (v2 + v1)t + 1/2(0)t^2
L = (v2 + v1)t
For the child:
The child is at rest relative to the second beltway, so its initial velocity (Vi) is 0. The distance the child needs to travel is -L (as it's going in the opposite direction to the man).
Using the equation of motion:
Xf = Xi + Vit
(-L) = 0 + 0t
-L = 0
Since the child starts at rest, it will take an infinite amount of time for the child to reach the man. It's because the child's velocity is zero and remains zero throughout.
Therefore, the answer to part (c) is that the man will never pass the child on the adjacent beltway.
For part C, you are correct in setting up the equations of motion for both the man and the child.
Let's start by defining the variables:
- Let t be the time it takes for the man to pass the child.
- Let V1 be the speed of the first beltway.
- Let V2 be the speed of the man relative to the beltway.
- Let L be the length of the beltway.
Now, let's set up the equations of motion:
For the man:
X_man = X_man_initial + V_man * t + 1/2 * a_man * t^2
L = (V2 + V1) * t
For the child:
X_child = X_child_initial + V_child * t
-L = -V1 * t
We assume that both the man and the child start at the same time.
Now we can solve these equations simultaneously to find the time t when the man passes the child.
Substituting the second equation into the first equation for the child's position, we get:
-L = -V1 * t = X_child_initial + V_child * t
Simplifying the equation, we get:
(X_child_initial + L) / V_child = t
Substituting this value of t into the equation for the man's position, we get:
L = (V2 + V1) * [(X_child_initial + L) / V_child]
Now we can solve this equation for L:
L * V_child = (V2 + V1) * (X_child_initial + L)
L * V_child = V2 * X_child_initial + L * V2 + V1 * X_child_initial + L * V1
L * (V_child - V2 - V1) = V2 * X_child_initial + V1 * X_child_initial
L = (V2 * X_child_initial + V1 * X_child_initial) / (V_child - V2 - V1)
So, to find the time it takes for the man to pass the child, we need to find the initial position of the child relative to the second beltway (X_child_initial), the speed of the child relative to the second beltway (V_child), and the speeds of the beltways (V1 and V2).