A 90 m long train begins uniform acceleration from rest. The front of the train has a speed of 21 m/s when it passes a railway worker who is standing 184 m from where the front of the train started. What will be the speed of the last car as it passes the worker? (See Fig. 2-36.)

To find the speed of the last car as it passes the worker, we can start by finding the acceleration of the train.

We know the initial velocity (u) of the train is 0 m/s, the final velocity (v) is 21 m/s, and the displacement (s) is 184 m - 90 m = 94 m (since the train is only 90 m long).

We can use the equation of motion:

v^2 = u^2 + 2as

Rearranging the equation, we have:

a = (v^2 - u^2) / (2s)

Substituting the given values, we have:

a = (21^2 - 0^2) / (2 × 94)

Simplifying, we get:

a = 66.915 m/s^2

Next, we can use this acceleration to find the time it takes for the train to pass the worker, assuming uniform acceleration.

We can use the equation of motion:

v = u + at

Rearranging the equation, we have:

t = (v - u) / a

Substituting the given values, we have:

t = (21 - 0) / 66.915

Simplifying, we get:

t = 0.314 s

Finally, to find the speed of the last car as it passes the worker, we can use the equation of motion:

v = u + at

Substituting the values, we have:

v = 0 + 66.915 × 0.314

Simplifying, we get:

v ≈ 21 m/s

Therefore, the speed of the last car as it passes the worker is approximately 21 m/s.

To find the speed of the last car as it passes the worker, we can use the equations of motion for uniformly accelerated linear motion.

Let's break down the problem:

Given:
- Length of the train (L) = 90 m
- Initial speed of the train (u) = 0 m/s
- Final speed of the front of the train (vF) = 21 m/s
- Distance of the railway worker from the starting point (S) = 184 m

We need to find the final speed of the last car (vL).

Using the equation of motion:
v^2 = u^2 + 2aS

where v is the final speed, u is the initial speed, a is the acceleration, and S is the distance traveled.

Since the train is in uniform acceleration, we can calculate the acceleration (a) using another equation:
v = u + at

where t is the time it takes for the front of the train to reach the worker.

First, let's calculate the acceleration:
vF = u + a * t
21 m/s = 0 m/s + a * t

Since the train starts from rest (u = 0), we can simplify the equation to:
21 m/s = a * t

Now, we can calculate the time it takes for the front of the train to reach the worker:
t = S / vF
t = 184 m / 21 m/s
t ≈ 8.762 seconds

Using this value of t, we can find the acceleration:
21 m/s = a * 8.762 seconds
a ≈ 2.395 m/s^2

Now that we have the acceleration, we can find the final speed of the last car (vL):
v^2 = u^2 + 2aS
vL^2 = (0 m/s)^2 + 2 * 2.395 m/s^2 * 90 m
vL^2 ≈ 431.1 m^2/s^2

Finally, taking the square root of vL^2, we can find the speed of the last car (vL):
vL ≈ √431.1 m^2/s^2
vL ≈ 20.8 m/s

Therefore, the speed of the last car as it passes the worker is approximately 20.8 m/s.

seriously bruh

First calculate the acceleration of the front of the train. When its velocity is V after traveling a distance X from a standing start,

V^2 = 2 a X, so
a = V^2 /(2X) = 1.198 m/s^2

When the last car passes the worker, the first car will have traveled 184 + 90 = 274 m

Use again V^2 = 2 a X, with the new value of X = 274. (a remains the same). This new V is the velocity of both the first and the last car at that time.

V^2 = 2*1.198*274
V = 25.6 m/s