You're an investigator for the National Transportation Safety Board, examining a subway accident in which a train going at 80km/h collided with a slower train traveling in the same direction at 20km/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 2.0m/s2 when it was 40m from the slower train, while the slower train continued at constant speed. What do you report
Based on the information provided, I will first calculate the time it took for the faster train to collide with the slower train. Then, using this time and the given distance, I will determine the relative speed of the collision.
To find the time taken for the faster train to collide, we can use the equation of motion:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Let's denote the final velocity of the faster train as v_f, the initial velocity as u_f, the acceleration as a_f, and the displacement as s_f. From the information given, we have:
v_f = 0 (since the faster train eventually stops)
u_f = 80 km/h = 80 * (1000/3600) m/s = 22.22 m/s
a_f = -2.0 m/s^2
s_f = 40 m
Plugging these values into the equation, we can solve for the time taken (t_f) for the faster train to reach the slower train:
0 = (22.22)^2 + 2(-2.0)s_f
Simplifying the equation:
0 = 493.26 - 4s_f
4s_f = 493.26
s_f = 123.32 m
Now that we have the displacement, we can find the time taken (t_f) using the equation of motion:
s_f = u_f * t_f + 0.5 * a_f * (t_f)^2
123.32 = 22.22*t_f - 0.5*2.0*(t_f)^2
Rearranging the equation:
0 = -2*(t_f)^2 + 22.22*t_f - 123.32
Using the quadratic formula, we can solve for t_f:
t_f = (-b ± √(b^2 - 4ac)) / (2a)
t_f = (-22.22 ± √((22.22)^2 - 4*(-2)*(-123.32))) / (2*(-2))
Calculating the values:
t_f ≈ 2.588 seconds or t_f ≈ 9.13 seconds
Since the train began negatively accelerating at 40m from the slower train, we can use the time taken (t_f) and the initial velocity (u_f) to calculate the relative speed of the collision.
Relative speed = u_f - a_f * t_f
Using the first solution for t_f:
Relative speed ≈ 22.22 - (-2.0) * 2.588 ≈ 27.396 m/s
Using the second solution for t_f:
Relative speed ≈ 22.22 - (-2.0) * 9.13 ≈ 40.486 m/s
Therefore, I would report that the relative speed of the collision between the two trains is approximately 27.396 m/s (or approximately 27.4 m/s when rounded to one decimal place) or 40.486 m/s (or approximately 40.5 m/s when rounded to one decimal place), depending on the time taken for the collision.