a train stopping distance, even when full emergency brakes are engaged, is 1.3 km. if the train was travelling at an initial velocity of 90 km/h [forward], determine its acceleration under full emergency brakes.

Since it is braking, the acceleration is negative.

Use the standard equation
Vf^2-Vi^2=2(a)S
where
Vf = final velocity (=0)
Vi = initial velocity (90 km/h=90*1000/3600=25 m/s)
a=acceleration required (m/s^2)
S=distance = 1.3 km = 1300 m

0^2-25^2=2(a)1300
Solve for a (in m/s^2)

Well, if I were a train, I would hope that my emergency brakes could stop me from being a "train-wreck"! But let's calculate the acceleration and get serious.

To determine the acceleration, we need to use the formula of motion known as the uniformly accelerated motion equation:

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s, since the train stops)
u = initial velocity (90 km/h = 25 m/s)
a = acceleration (what we need to find)
s = stopping distance (1.3 km = 1300 m)

Plugging in the values we know, the equation becomes:

0^2 = (25)^2 + 2a(1300)

Simplifying further:

0 = 625 + 2600a

Subtracting 625 from both sides:

-625 = 2600a

Dividing through by 2600:

a ≈ -0.24 m/s^2

So, the acceleration under full emergency brakes is approximately -0.24 m/s^2. Negative because the train is decelerating. Keep in mind this value is an average, and in real life conditions and situations may vary!

To determine the acceleration of the train under full emergency brakes, we can use the formula for braking distance:

Braking distance = (initial velocity^2) / (2 * acceleration)

Given that the train's initial velocity is 90 km/h and the braking distance is 1.3 km, we can rearrange the formula to solve for acceleration:

acceleration = (initial velocity^2) / (2 * braking distance)

Converting the initial velocity to m/s:

initial velocity = 90 km/h * (1000 m/1 km) * (1 h/3600 s) = 25 m/s

Substituting the values into the equation:

acceleration = (25 m/s)^2 / (2 * 1.3 km * (1000 m/1 km))

Simplifying:

acceleration = (625 m^2/s^2) / (2 * 1.3 * 10^3 m)

acceleration = 625 / (2 * 1.3 * 10^3)

Calculating:

acceleration = 625 / 2.6 * 10^3

acceleration = 240.38 m^2/s^2

Therefore, the acceleration of the train under full emergency brakes is approximately 240.38 m^2/s^2.

To determine the acceleration of the train under full emergency brakes, we can use the kinematic equation:

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s, since the train comes to a stop)
u = initial velocity (90 km/h or 25 m/s)
a = acceleration (unknown)
s = stopping distance (1.3 km or 1300 m)

Rearranging the equation, we have:

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

Let's calculate it step by step:

1. Convert the initial velocity to meters per second:
u = 25 m/s

2. Plug in the values into the equation:
a = (0^2 - 25^2) / (2 * 1300)

3. Simplify the equation:
a = -625 / 2600

4. Calculate the acceleration:
a ≈ -0.24 m/s^2

Therefore, the acceleration of the train under full emergency brakes is approximately -0.24 m/s^2. The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, i.e., deceleration.