A fugitive tries to hop on a freight train traveling at a constant speed of 5.2 m/s . Just as an empty box car passes him, the fugitive starts from rest and accelerates at a = 1.4 m/s2 to his maximum speed of 5.9 m/s .
So,How long does it take him to catch up to the empty box car?
Well, I hope this fugitive has a good sense of timing because we've got some math to do! Let's calculate how long it takes for him to catch up to the empty box car.
First, we need to find out the distance the fugitive needs to cover to catch up. We can use the equation:
distance = initial velocity x time + 0.5 x acceleration x time^2
Since the fugitive starts from rest, his initial velocity is 0, and his acceleration is 1.4 m/s^2. We'll let 't' represent the time it takes for him to catch up.
The distance the empty box car travels during this time can be calculated by multiplying its constant speed by 't'. Since it is traveling at 5.2 m/s, the equation would be:
distance = 5.2 m/s x t
Now, we set these two distances equal to each other and solve for 't':
0 + 0.5 x 1.4 m/s^2 x t^2 = 5.2 m/s x t
After simplifying and rearranging terms, we get:
0.7 t^2 - 5.2 t = 0
Now, we can solve this quadratic equation for 't'. *cue the suspenseful music*
Using some math magic (or in this case, the quadratic formula), we find that 't' is either 0 seconds or approximately 7.43 seconds.
Since 't' cannot be 0 (that would mean the fugitive caught up instantaneously, which is highly improbable), we can conclude that it takes the fugitive approximately 7.43 seconds to catch up to the empty box car.
I hope the fugitive brought a stopwatch with them, because timing is everything when it comes to train-hopping!
To find the time it takes for the fugitive to catch up to the empty box car, we can use the kinematic equation:
vf = vi + at
Where:
vf = final velocity
vi = initial velocity
a = acceleration
t = time
In this case, the fugitive's initial velocity is 0 m/s (as he starts from rest), the acceleration is 1.4 m/s², and his maximum velocity is 5.9 m/s.
First, let's calculate the time it takes for the fugitive to reach his maximum velocity:
vf = vi + at
5.9 m/s = 0 m/s + (1.4 m/s²) * t
Rearranging the equation to solve for time, we have:
t = (5.9 m/s) / (1.4 m/s²)
t = 4.214 seconds
So it takes the fugitive approximately 4.214 seconds to reach his maximum speed.
Now, let's find out how far the empty box car travels during this time:
distance = velocity * time
distance = (5.2 m/s) * (4.214 seconds)
distance ≈ 21.8388 meters
Therefore, the fugitive needs to cover a distance of approximately 21.8388 meters to catch up to the empty box car.
To find out how long it takes for the fugitive to catch up to the empty box car, we can use the equation of motion:
v = u + at
where:
v = final velocity (the maximum speed of the fugitive)
u = initial velocity (0 m/s because the fugitive starts from rest)
a = acceleration
t = time
From the given information, we know that the maximum speed of the fugitive is 5.9 m/s and the acceleration is 1.4 m/s². Let's substitute these values into the equation and solve for t:
5.9 m/s = 0 m/s + (1.4 m/s²) * t
Simplifying the equation:
5.9 m/s = 1.4 m/s² * t
Now, divide both sides of the equation by 1.4 m/s² to isolate t:
t = 5.9 m/s / 1.4 m/s²
Calculating the result:
t ≈ 4.21 seconds
Therefore, it takes approximately 4.21 seconds for the fugitive to catch up to the empty box car.
How long does it take him to get to max speed? After that, how long does it take to catch up?
5.9/1.4 = 4.214 seconds to max out
So he will catch up after t seconds, where
5.2t = 1/2 * 1.4 * 4.214^2 + 5.9(t-1.4214)
t = -5 seconds. ... ???
So, that means he caught up to the train before he reached his maximum speed. That is,
5.2t = 1/2 * 1.4 * t^2
7.429 seconds