a pedestrian is running at his maximum speed of 6.0 m/s trying to catch a bus that is stopped at a traffic light. when he is 16m from the bus, the light changes and the bus pulls away from the pedestrian with an acceleration of 1.0 m/s^2. does the pedestrian catch the bus and, if so, how far does he have to run. (if not, what is the pedestrian's distance of closest approach)
6 t - 16 = 1/2 1 t^2 ... 12 t - 32 = t^2 ... t^2 - 12 t + 32 = 0 ... (t - 8) (t - 4) = 0
looks like he catches the bus in four seconds
... if he kept running, he would pass the bus
... which would catch back up to him at the eight second mark
running for four seconds means he travels 24 m
Well, let's see if the pedestrian catches the bus with his lightning-fast speed of 6.0 m/s.
To find out, we need to calculate the time it takes for the bus to start moving while the pedestrian is still running towards it. We can use the equation:
distance = initial velocity × time + (1/2) × acceleration × time^2
Since the bus is initially at rest and the pedestrian is running towards it, we can simplify the equation to:
16 m = 6.0 m/s × time + (1/2) × 1.0 m/s^2 × time^2
Now, let's solve it and find out if the pedestrian can catch the bus!
Wait, I need a second...I'm trying to crunch the numbers, but my calculator seems to be having a laugh at my expense. Oh, there it goes! Alright, let's calculate...
Using the quadratic formula, the time it takes for the bus to start moving while the pedestrian is still running towards it is approximately 1.89 seconds.
Now that we know the time, we can calculate how far the pedestrian has to run to catch the bus:
distance = initial velocity × time = 6.0 m/s × 1.89 s = 11.34 meters
So, unfortunately, the pedestrian doesn't catch the bus. He comes really close, but he falls short by around 4.66 meters.
But hey, look on the bright side! At least he got a good workout, right? Maybe he can catch the next bus and claim it was all part of his fitness routine.
To determine if the pedestrian catches the bus, we need to compare their distances covered over time. Let's break down the problem step-by-step:
Step 1: Calculate the time it takes for the pedestrian to reach the bus.
We can use the formula distance = speed × time.
The initial distance between the pedestrian and the bus is 16m.
The pedestrian's speed is 6.0 m/s.
Let's assume the time it takes for the pedestrian to reach the bus is t.
Distance covered by the pedestrian = 6.0 m/s × t
Distance covered by the bus = 0 (since it is stopped)
According to the given information, the pedestrian reaches the bus when the traffic light changes. At this point, the bus starts pulling away.
Step 2: Calculate the distance covered by the bus when the pedestrian reaches it.
The bus starts accelerating with an acceleration of 1.0 m/s^2.
We know that distance = initial velocity × time + 0.5 × acceleration × time^2.
Since the bus has an initial velocity of 0, we can simplify the equation to distance = 0.5 × acceleration × time^2.
Let's assume the time it takes for the pedestrian to reach the bus is also t'.
Thus, the distance covered by the bus when the pedestrian reaches it would be 0.5 × 1.0 m/s^2 × t'^2.
Step 3: Solve for t by equating the distances covered by the pedestrian and the bus.
6.0 m/s × t = 0.5 × 1.0 m/s^2 × t'^2
Step 4: Solve for t' using the equation for the distance covered by the bus.
The initial distance between the pedestrian and the bus is 16m.
Distance covered by the bus = 0.5 × 1.0 m/s^2 × t'^2
Now we have two equations:
1. 6.0 m/s × t = 0.5 × 1.0 m/s^2 × t'^2
2. Distance = 16m = 0.5 × 1.0 m/s^2 × t'^2
Step 5: Solve the system of equations to find the values of t and t'.
From equation 2, we can calculate t'^2 as follows:
16m = 0.5 × 1.0 m/s^2 × t'^2
32m = t'^2
t' ≈ √32 ≈ 5.66s
Now we can substitute t' back into equation 1 to find t:
6.0 m/s × t = 0.5 × 1.0 m/s^2 × (5.66s)^2
6.0 m/s × t = 0.5 × 1.0 m/s^2 × 32m
6.0 m/s × t = 16m
t = 16m ÷ 6.0 m/s
t ≈ 2.67s
Step 6: Calculate the distance covered by the pedestrian.
Now that we know the time it takes for the pedestrian to reach the bus is approximately 2.67 seconds, we can calculate the distance covered by the pedestrian:
Distance covered by the pedestrian = 6.0 m/s × t
Distance covered by the pedestrian = 6.0 m/s × 2.67s
Distance covered by the pedestrian ≈ 16.01m
The pedestrian catches the bus, and he needs to run approximately 16.01 meters to reach it.
To solve this problem, we need to determine whether the pedestrian catches the bus, and if so, how far he has to run. If he does not catch the bus, then we need to find the pedestrian's distance of closest approach.
Let's break down the problem into two parts: the time it takes for the pedestrian to reach the bus, and the subsequent motion of the bus.
1. Time taken for the pedestrian to reach the bus:
We can use the equation of motion: distance = initial velocity * time + (1/2) * acceleration * time^2. We know that the pedestrian's initial velocity is 6.0 m/s, the distance is 16m, and there is no initial acceleration. Let's find the time it takes for the pedestrian to reach the bus.
distance = 6.0 m/s * time
16m = 6.0 m/s * time
time = 16m / 6.0 m/s
time = 2.67 seconds (approximately)
Therefore, it takes the pedestrian approximately 2.67 seconds to reach the bus.
2. Subsequent motion of the bus:
After the pedestrian reaches the bus, the bus starts accelerating with an acceleration of 1.0 m/s^2. We need to determine if the pedestrian catches the bus. We can do this by comparing their distances covered over the same time interval.
Distance covered by the pedestrian = 6.0 m/s * 2.67 seconds
= 16.02 meters (approximately)
Distance covered by the bus, starting from rest and accelerating with an acceleration of 1.0 m/s^2:
Using the equation of motion: distance = (1/2) * acceleration * time^2
Distance covered by the bus = (1/2) * 1.0 m/s^2 * (2.67 seconds)^2
= (1/2) * 1.0 m/s^2 * 7.1289 seconds^2
= 3.5645 meters (approximately)
Since the pedestrian covers a greater distance (16.02 meters) than the bus (3.5645 meters), the pedestrian catches the bus.
Therefore, the pedestrian catches the bus, and he has to run a total distance of 16.02 meters.