When a high-speed passenger train traveling at 124 km/h rounds a bend, the engineer is shocked to see that a locomotive has improperly entered onto the track from a siding and is a distance D=666 m ahead. The locomotive is moving at 21 km/. The engineer of the high-speed train immediately applies the brakes.
A) What must be the magnitude of the resulting constant deceleration of a collision is to be just avoided?
This is what I did:
I changed 124 km/h to 34.4 m/s and changed 21 km/h to 5.83 m/s. I believe at this point I am supposed to use one of the kinematic equations, but I am not sure which one. I think delta x = 34.4 - 5.83 = 28.57m. That would give me a position and I don't think there is an acceleration/deceleration rate so any equation with an a can't be used? That leaves me with this equation: delta x = 1/2(Vo + V)t
Is this right to use? Would Vo be 34.4 and V 5.83? Please Help!
B) Assume the engineer is at x=0 when, at t=0, he first spots the locomotive. Sketch x(t) curves for the locomotive and high-speed train for the cases in which a collision is just avoided and is not quite avoided.
I'm struggling with velocity time graphs and I am not sure about this, but an explanation would be nice.
Equations for constant acceleration:
v = Vi + a t
x = Xo + Vi t + (1/2) a t^2
here
for the high speed train
starts at 0 meters if locomotive starts at 666 meters
acceleration = negative a
v = 34.4 - a t until v = 5.83
so
5.83 = 34.4 - a t
a t = 28.6
x = 0 + 34.4 t -.5 a t^2
for the locomotive, no acceleration
v = 5.83
x = 666 + 5.83 t + 0 = 666 + 5.83 t
same x if they are at point x when their speeds are the same
so
34.4 t - .5 a t^2 = 666 + 5.83 t
solve quadratic and finish.
oh, not even a quadratic because at=28.6
34.4 t - .5(28.6) t = 666 + 5.83 t
When I solve the part in the second post I get 46.67 sec. When I solve the quadratic I get -1014.14 and 1071.28
I arranged it as -0.5at^2 + 28.57 - 666 = 0
Also, I guess the wording of the question with "magnitude of the resulting constant deceleration" made me think the answer should be acceleration in m/s^2 but this is not the case as the answer is time in seconds. Why is this?
The units are wrong, it did ask for an acceleration.
To solve this problem, let's analyze the situation using the equations of motion.
A) The goal is to determine the magnitude of the constant deceleration required to avoid a collision with the locomotive. We can assume that the high-speed train and the locomotive will reach the same position at the same time.
The equations of motion in this case are:
- For the high-speed train:
x_train = Vo_train * t - (1/2) * a_train * t^2
- For the locomotive:
x_locomotive = Vo_locomotive * t
where x_train and x_locomotive represent the positions of the train and locomotive at any given time t (measured from the point where the engineer spots the locomotive), Vo_train and Vo_locomotive are the initial velocities of the train and the locomotive, and a_train is the constant deceleration of the train.
Given the initial velocities Vo_train = 34.4 m/s and Vo_locomotive = 5.83 m/s, and the distance between them D = 666 m, we can equate the positions of the train and the locomotive at the time of collision avoidance:
Vo_train * t - (1/2) * a_train * t^2 = Vo_locomotive * t + D
Substituting the given values, we have:
34.4 * t - (1/2) * a_train * t^2 = 5.83 * t + 666
Simplifying the equation further, we get:
(1/2) * a_train * t^2 - 28.57 * t + 666 = 0
This is a quadratic equation in terms of "t". By solving this equation, you can find the time at which the collision is avoided. From there, you can calculate the deceleration using the equation:
a_train = (Vo_train - Vo_locomotive) / t
B) To sketch the x(t) curves for both the train and the locomotive, you need to consider the different scenarios - when a collision is just avoided and when it is not quite avoided.
1. Collision just avoided: In this case, the train and the locomotive reach the same position at the same time. The x(t) curve for the train will be a decreasing parabola, indicating deceleration, while the x(t) curve for the locomotive will be a straight line with a constant positive slope, indicating constant speed.
2. Collision not quite avoided: In this scenario, the train will collide with the locomotive. The x(t) curve for the train will be a decreasing parabola with a steeper slope compared to the curve in the previous scenario. The x(t) curve for the locomotive will still be a straight line with a constant positive slope. The point of collision will be the intersection of these two curves.
Note: Without specific values for time, position, and acceleration, it is difficult to provide an accurate sketch. However, understanding the general behavior of the x(t) curves will help you visualize the situations.