You're an investigator for the National Transportation Safety Board, examining a subway accident in which a train going at 88km/h km/h collided with a slower train traveling in the same direction at 25km/h km/h . Your job is to determine the relative speed of the collision, to help establish new crash standards. The faster train's "black box" shows that it began negatively accelerating at 1.8m/s 2 m/s^2 when it was 40m m from the slower train, while the slower train continued at constant speed.
Well, well, well, looks like we've got ourselves a good old subway crash investigation! Hang on tight, folks, because Clown Bot is here to bring some humor to this crash course on physics!
Alright, let's break it down. We have a speedy train going at a whopping 88 km/h and a slower train chugging along at a measly 25 km/h. The faster train's "black box" reveals it started negatively accelerating at 1.8 m/s^2 when it was 40m away from the slower train.
Now, first things first, let's convert those speeds into the same units. I like to call it the "Unit Party" where everyone has to get on the same wavelength. So, 88 km/h becomes 24.4 m/s (approximately) and 25 km/h transforms into 6.9 m/s (roughly).
Next up, we need to determine the relative speed of the collision. Picture it like this: the faster train is running towards the slower one, but it's slowing down. So, we need to find out how quickly it slows down before the collision.
Using our trusty physics formulas, we can find the time it takes for the faster train to reach the slower one. We use the equation: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.
Plug in the values we have, and we get v^2 = (24.4 m/s)^2 + 2 * (-1.8 m/s^2) * 40 m. Math time! Calculate it, and we find that v is approximately 20.2 m/s.
Now, to find the relative speed, we subtract the slower train's speed from the faster train's speed: 20.2 m/s - 6.9 m/s. Drumroll, please! The relative speed of the collision is approximately 13.3 m/s.
So, there you have it, dear investigator! A humor-filled explanation of how to determine the relative speed of this subway collision. Remember, safety standards are no laughing matter, but Clown Bot is always here to make learning a little more entertaining!
To determine the relative speed of the collision between the faster train and the slower train, we can analyze the information provided:
1. Convert the given speeds from km/h to m/s:
- Faster train speed = 88 km/h = (88 * 1000) / (60 * 60) = 24.44 m/s
- Slower train speed = 25 km/h = (25 * 1000) / (60 * 60) = 6.94 m/s
2. Calculate the time it takes for the faster train to collide with the slower train:
- Relative position = distance - 40 m (Given that faster train starts negatively accelerating from this position)
- Let's assume the collision occurs at time t (seconds)
- The faster train covers the initial distance of 40m with a constant velocity before it starts negatively accelerating
- Using the equation: distance = initial velocity * time + (1/2) * acceleration * time^2
- 40 = 24.44 * t + (1/2) * (-1.8) * t^2
3. Solve the quadratic equation to find the time of collision:
- Rearranging the equation: -0.9t^2 + 24.44t - 40 = 0
- Using the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a)
- a = -0.9, b = 24.44, and c = -40
- Calculate using the quadratic formula:
t = (-24.44 ± √((24.44)^2 - 4 * (-0.9) * (-40))) / (2 * -0.9)
= (-24.44 ± √(598.7536)) / (-1.8)
= (-24.44 ± 24.47) / (-1.8)
- So, we have two possible values for t: t1 = 0.0156 seconds and t2 = 53.241 seconds
4. Determine the relative speed of the collision:
- Since the slower train continues at a constant speed, the relative speed of the collision is the difference between the faster train's speed and the slower train's speed
- Considering the slower train's speed was 6.94 m/s, the relative speed at the time of collision would be:
- Relative speed at t1 = 24.44 m/s - 6.94 m/s = 17.5 m/s
- Relative speed at t2 = 24.44 m/s - 6.94 m/s = 17.5 m/s
Therefore, the relative speed of the collision between the faster train and the slower train is 17.5 m/s.
To determine the relative speed of the collision between the two trains, we need to consider the velocities of both trains and their positions at the time of collision.
Given:
- The faster train's initial velocity (vf) is 88 km/h, which we need to convert to m/s.
- The slower train's velocity (vs) is 25 km/h, which we also need to convert to m/s.
- The faster train decelerates negatively at 1.8 m/s^2.
- The faster train's position at the time of deceleration is 40 m.
To make the units consistent, we need to convert the velocities into meters per second:
- vf = 88 km/h * (1000 m/1 km) * (1 h/3600 s) = 24.44 m/s
- vs = 25 km/h * (1000 m/1 km) * (1 h/3600 s) = 6.94 m/s
Now, we need to evaluate the distance and time it took for the faster train to come to a stop. We can use the following kinematic equation:
vf^2 = vi^2 + 2ad
Where:
- vf is the final velocity (0 m/s since the train came to a stop)
- vi is the initial velocity (24.44 m/s)
- a is the acceleration (-1.8 m/s^2)
- d is the distance (40 m)
Rearranging the equation:
0 = (24.44 m/s)^2 + 2 * (-1.8 m/s^2) * d
Now we can solve for d, the distance it took for the faster train to stop:
d = (24.44 m/s)^2 / (2 * 1.8 m/s^2) = 174.95 m
Since the faster train started decelerating when it was 40 m away from the slower train, we can subtract this distance from the total distance it took to stop to find the position of the slower train at the time of collision:
Position slower train = 174.95 m - 40 m = 134.95 m
Now, we can calculate the relative speed of the collision. We need to subtract the velocity of the slower train from the velocity of the faster train:
Relative speed = vf - vs = 24.44 m/s - 6.94 m/s = 17.5 m/s
Therefore, the relative speed of the collision between the two trains is 17.5 m/s.