a uniformly accelerating train passes a green light signal at25.0 km/hr it passes a second light 125m farther down the track 12.0s later what is the trains acceleration? what is the trains velocity at the second light?
x = Vi t + (1/2) a t^2
125 = (25000/3600)(12) + .5 * a * 144
solve for a
then
v = Vi + a t
where did you get the 144
ok i get
correct me if im wrong solve for a = 102
and v = 1249
A = 51.15
V= 638.8
To find the train's acceleration and velocity at the second light, we can use the kinematic equations of motion. The equations we need are:
1. v = u + at
2. s = ut + (1/2)at^2
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time elapsed
s = distance traveled
Let's find the acceleration first:
Given:
Initial velocity (u) = 25.0 km/hr
Final velocity (v) is unknown
Time elapsed (t) = 12.0 s
Distance traveled (s) = 125 m
First, we need to convert the initial velocity from km/hr to m/s:
u = 25.0 km/hr * (1000 m / 3600 s) = 6.94 m/s (approx.)
Using equation 1, we can find the acceleration:
v = u + at
0 = 6.94 m/s + a * 12.0 s
-6.94 m/s = 12.0 a
a = -6.94 m/s / 12.0 s ≈ -0.577 m/s²
Therefore, the train's acceleration is approximately -0.577 m/s² (negative since it is decelerating).
To find the train's velocity at the second light, we can use equation 1 again:
v = u + at
v = 6.94 m/s + (-0.577 m/s²) * 12.0 s
v = 6.94 m/s - 6.92 m/s (approx.)
v ≈ 0.02 m/s
Therefore, the train's velocity at the second light is approximately 0.02 m/s.