A)If a freight train can accelerate at 0.050 m/sec^2, how far must it travel to increase its speed from 2.0 to 8.0 m/sec? B) how long will it take the train to accomplish this?

A. Vf^2 = Vo^2 + 2ad = 8^2.

2^2 + 2 * 0.05 * d = 64,
4 + 0.1d = 64,
0.1d = 64 - 4 = 60,
d = 60 / 0.1 = 600m.

B. Vf = Vo + at = 8m/s.
2 + 0.05t = 8,
0.05t = 8 - 2 = 6,
t = 6 / 0.05 = 120s.

To find the distance the train must travel to increase its speed from 2.0 to 8.0 m/s and the time it will take, we can use the following kinematic equations:

A) The equation relating distance, initial velocity, final velocity, and acceleration is:
\(v^2 = u^2 + 2as\),

where:
v = final velocity (8.0 m/s),
u = initial velocity (2.0 m/s),
a = acceleration (0.050 m/s^2),
s = distance.

Rearranging the equation to isolate s, we have:
\(s = \frac{v^2 - u^2}{2a}\).

Substituting the given values:
\(s = \frac{(8.0)^2 - (2.0)^2}{2(0.050)}\),
\(s = \frac{64 - 4}{0.1}\),
\(s = \frac{60}{0.1}\),
\(s = 600 \, \text{meters}\).

Therefore, the train must travel 600 meters to increase its speed from 2.0 to 8.0 m/s.

B) To find the time it will take for the train to accomplish this, we can use the equation:
\(v = u + at\),

where:
v = final velocity (8.0 m/s),
u = initial velocity (2.0 m/s),
a = acceleration (0.050 m/s^2),
t = time.

Rearranging the equation to solve for t, we have:
\(t = \frac{v - u}{a}\),

Substituting the given values:
\(t = \frac{8.0 - 2.0}{0.050}\),
\(t = \frac{6.0}{0.050}\),
\(t = 120 \, \text{seconds}\).

Therefore, it will take the train 120 seconds, or 2 minutes, to increase its speed from 2.0 to 8.0 m/s.

To solve these problems, we can use the kinematic equations that relate displacement, acceleration, initial velocity, time, and final velocity.

A) To determine the distance the freight train must travel, we can use the equation:

vf^2 = vi^2 + 2ad

Where:
- vf is the final velocity (8.0 m/s),
- vi is the initial velocity (2.0 m/s),
- a is the acceleration (0.050 m/s^2), and
- d is the distance we need to find.

Rearrange the equation to solve for d:

d = (vf^2 - vi^2) / (2a)

Substituting the given values:

d = (8.0^2 - 2.0^2) / (2 * 0.050)
= (64 - 4) / 0.1
= 60 / 0.1
= 600 meters

Therefore, the freight train must travel 600 meters to increase its speed from 2.0 to 8.0 m/s.

B) To determine the time it will take for the train to accomplish this, we can use another kinematic equation that relates initial velocity, final velocity, acceleration, and time:

vf = vi + at

Rearrange the equation to solve for t:

t = (vf - vi) / a

Substituting the given values:

t = (8.0 - 2.0) / 0.050
= 6.0 / 0.050
= 120 seconds

Therefore, it will take the train 120 seconds (or 2 minutes) to accomplish this acceleration.