A bus moving at a constant speed of 5.00 m/s passes you. When the rear of the bus is 12.0 m past you, you run toward it with acceleration of 0.96 m/s^2. How far would you have to run before you catch up with the rear of the bus, and how fast must be running then?
5.00 t + 12.0 = 1/2 * 0.96 * t^2
0.48 t^2 - 5.00 t - 12 = 0
solve the quadratic for t (the positive value)
a * t is the speed at catch-up
1/2 a t^2 is the distance
Great!
i got 0,-12
what is (a)?
for the speed to catch-up is a = 5, or 12?
for the distance (a) should be 0.96 right?
never mind, got it. Thanks btw :)
Well, catching up with a bus can be quite a runny situation! Let's calculate the distance you would have to run before catching up with the rear of the bus.
We know that the bus is moving at a constant speed of 5.00 m/s. When the rear of the bus is 12.0 m past you, your runny adventure begins. So, the distance between you and the bus is initially 12.0 meters.
Now, let's find out how long it takes for you to catch up with the bus. To do that, we need to see when the distances covered by you and the bus become equal. The bus is moving at a constant speed, so the distance it covers will be the product of its speed and time.
The distance you cover is a bit trickier. You start from rest and have an acceleration of 0.96 m/s^2. We can use the equation:
d = 1/2 * a * t^2
where d is the distance, a is the acceleration, and t is the time.
Since you're chasing the bus, your distance should be the initial distance between you and the bus, plus the distance you cover while catching up. Let's call the time it takes for you to catch up with the bus "t".
So, your distance is:
12.0 m + (1/2 * 0.96 m/s^2 * t^2)
Now, let's equate the distances covered by you and the bus:
5.00 m/s * t = 12.0 m + (1/2 * 0.96 m/s^2 * t^2)
Now, we can solve this equation to find the time it takes for you to catch up with the bus.
t = sqrt((2 * (12.0 m)) / (0.96 m/s^2))
After calculating, t is approximately 5.77 seconds.
Great! Now that we know the time it takes for you to catch up with the bus, we can calculate the distance you would have to run. Using the equation:
d = 1/2 * a * t^2
d = 1/2 * 0.96 m/s^2 * (5.77 s)^2
d is approximately 16.05 meters.
So, you would have to run approximately 16.05 meters to catch up with the rear of the bus.
As for how fast you must be running then, the bus is moving at a constant speed of 5.00 m/s. To catch up with it, you would need to match this speed. So, you must be running at 5.00 m/s to catch up with the rear of the bus.
Happy running!
To answer this question, we need to determine the time it takes for you to catch up with the rear of the bus. Let's break down the problem into different stages:
Stage 1: Bus passing
When the bus passes you, you are stationary, and the bus is moving at a constant speed of 5.00 m/s. Since you are stationary, the bus moves ahead by a distance of 12.0 m relative to you.
Stage 2: You start running
After the bus passes you, you start running towards it with an acceleration of 0.96 m/s^2. The bus continues to move at a constant speed.
To determine the time it takes for you to catch up with the rear of the bus, we need to find the time it takes to cover the initial separation distance of 12.0 m. We can use the following equation:
Δx = v₀t + (1/2)at²
Where:
Δx is the displacement (distance) you cover,
v₀ is your initial velocity (0 m/s since you start from rest),
t is the time taken,
a is your acceleration (0.96 m/s²).
Rearranging the equation to solve for time:
Δx - (1/2)at² = 0
Substituting the given values:
12.0 - (1/2)(0.96)t² = 0
Simplifying:
0.48t² = 12.0
Dividing by 0.48:
t² = 25.0
Taking the square root of both sides:
t ≈ 5.0 s
You will take approximately 5.0 seconds to cover the initial separation distance.
Now, let's move on to the second part of the question:
Stage 3: Catching up with the bus
During this stage, you are running towards the bus while it continues moving at 5.00 m/s. At the point of catching up, your final velocity must be equal to the velocity of the bus.
We can use the following equation to find how far you will have to run:
Δx = v₀t + (1/2)at²
Where:
Δx is the displacement (distance) you cover,
v₀ is your initial velocity (0 m/s since you start from rest),
t is the time taken,
a is your acceleration (0.96 m/s²).
Rearranging the equation to solve for distance:
Δx = (1/2)at²
Substituting the given values:
Δx = (1/2)(0.96)(5.0)²
Calculating:
Δx ≈ 6.0 m
Therefore, you will have to run approximately 6.0 meters before you catch up with the rear of the bus.
To find your final velocity when you catch up with the bus, we can use the equation:
v = v₀ + at
Where:
v is your final velocity,
v₀ is your initial velocity (0 m/s since you start from rest),
a is your acceleration (0.96 m/s²),
t is the time taken (5.0 s).
Substituting the given values:
v = (0.96)(5.0)
Calculating:
v ≈ 4.8 m/s
Therefore, when you catch up with the rear of the bus, you must be running at a speed of approximately 4.8 m/s.