A pedestrian is running at his maximum speed of 6.0m/s to catch a bus stopped by a traffic light. When he is 25m away from the bus the light changes and the bus accelerates uniformly at 1.0m/s^2. Find either (a) how far he has to run to catch the bus or (b) his frustration distance(closest approach)
a. d2 = d1 - 25,
6t = 0 + 0.5 * 1t^2 - 25,
6t - 0.5t^2 + 25 = 0,
-0.5t^2 + 6t + 25 = 0,
Use Quadratic Formula to find t:
t = 15.27s.
d = Vt = 6 * 15.27 = 91.6m.
10m
To find either (a) how far the pedestrian has to run to catch the bus or (b) his frustration distance (closest approach), we can use the equations of motion.
Let's solve for (a) first: how far the pedestrian has to run to catch the bus.
1. We need to find the time it takes for the bus to move from its initial position to the point where the pedestrian is, which is 25m away.
Using the equation: distance = initial velocity * time + (1/2) * acceleration * time^2
Let's assume the time taken by the bus to reach the pedestrian is t.
25 = 0 * t + (1/2) * 1.0 * t^2
25 = (1/2) * t^2
50 = t^2
t = sqrt(50)
2. Now that we know the time, we can find how far the pedestrian has to run by using their maximum speed.
Distance the pedestrian has to run = pedestrian's speed * time
Distance = 6.0 * sqrt(50) = 42.43m (rounded to two decimal places)
Therefore, the pedestrian has to run approximately 42.43m to catch the bus.
Now, let's solve for (b) the frustration distance or closest approach.
1. We need to find the time when the pedestrian and the bus are at the closest point.
To find this, we can equate the distance covered by the pedestrian and the distance covered by the bus during this time.
Let's assume the time when the closest point is achieved is t'.
For the pedestrian: distance covered = pedestrian's speed * t'
For the bus: distance covered = initial velocity * t' + (1/2) * acceleration * t'^2
The closest point is achieved when these distances are equal.
6.0 * t' = 0 * t' + (1/2) * 1.0 * t'^2
6.0 * t' = (1/2) * t'^2
12.0 = t'
2. Finally, we can find the frustration distance by substituting the time (t') into the equation for the pedestrian's distance covered.
Frustration distance = pedestrian's speed * t'
Frustration distance = 6.0 * 12.0 = 72.0m
Therefore, the pedestrian's frustration distance or closest approach is 72.0m.
To find either (a) how far the pedestrian has to run to catch the bus or (b) his frustration distance (closest approach), we need to calculate the time it takes for the pedestrian to catch up to the bus and then use this time to find the respective distances.
Let's start by finding the time it takes for the pedestrian to catch up to the bus.
We can use the equation of motion:
(distance) = (initial velocity) * (time) + (0.5) * (acceleration) * (time^2)
For the pedestrian:
distance_pedestrian = 25m (since he is initially 25m away from the bus)
initial velocity_pedestrian = 6.0m/s
acceleration_pedestrian = 0m/s^2 (since the pedestrian is not accelerating)
For the bus:
initial velocity_bus = 0m/s (since it is initially stopped)
acceleration_bus = 1.0m/s^2
Using the equation of motion for the pedestrian:
25m = 6.0m/s * t_pedestrian + 0.5 * 0m/s^2 * (t_pedestrian^2)
25m = 6.0m/s * t_pedestrian
Simplifying the equation:
t_pedestrian = 25m / 6.0m/s
t_pedestrian ≈ 4.17s (rounded to two decimal places)
Now that we have the time it takes for the pedestrian to catch up to the bus, let's calculate the distances.
(a) The distance the pedestrian has to run to catch the bus:
Using the equation of motion with the initial velocity of 6.0m/s (since the pedestrian is running at his maximum speed) and the time of 4.17s:
distance_pedestrian = 6.0m/s * 4.17s
distance_pedestrian ≈ 25.02m (rounded to two decimal places)
Therefore, the pedestrian has to run approximately 25.02 meters to catch the bus.
(b) The frustration distance (closest approach):
To find the frustration distance, we need to calculate the position of the pedestrian and the bus at the time of closest approach.
For the pedestrian:
position_pedestrian = initial position_pedestrian + (initial velocity_pedestrian * t_pedestrian)
position_pedestrian = 25m + (6.0m/s * 4.17s)
position_pedestrian ≈ 50.01m (rounded to two decimal places)
For the bus:
Using the equation of motion with the initial velocity of 0m/s and the time of 4.17s:
position_bus = initial position_bus + (initial velocity_bus * t_pedestrian) + (0.5 * acceleration_bus * (t_pedestrian^2))
position_bus = 0m + (0m/s * 4.17s) + (0.5 * 1.0m/s^2 * (4.17s^2))
position_bus ≈ 8.70m (rounded to two decimal places)
Now, the frustration distance is the difference between the positions of the pedestrian and the bus:
frustration_distance = position_pedestrian - position_bus
frustration_distance ≈ 50.01m - 8.70m
frustration_distance ≈ 41.31m (rounded to two decimal places)
Therefore, the frustration distance (closest approach) is approximately 41.31 meters.