A train is traveling at a speed of 80\,\dfrac{\text{km}}{\text{h}}80 h km 80, start fraction, start text, k, m, end text, divided by, start text, h, end text, end fraction when the conductor applies the brakes. The train slows with a constant acceleration of magnitude 0.5\,\dfrac{\text{m}}{\text{s}^2}0. 5 s 2 m 0, point, 5, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction. We want to find the distance the train travels from the time the conductor applies the brakes until the train comes to a complete stop. Which kinematic formula would be most useful to solve for the target unknown? Choose 1

The distance the train travels before coming to a (complete) stop = 40/81 km which is approximately 493.83 meters

Explanation:

The initial speed of the train u = 80 km/h = 22 2/9 m/s = 22. m/s

The magnitude of the constant acceleration with which the train slows, a = 0.5 m/s²

Therefore, we have the following suitable kinematic equation of motion;

v² = u² - 2 × a × s

Where;

v = The final velocity = 0 (The train comes to a stop)

s = The distance the train travels before coming to a stop

Substituting the  values gives;

0² = 22.² - 2 × 0.5 × s

2 × 0.5 × s = 22.²

s = 22.²/1 = 493 67/81 m = 40/81 km

The distance the train travels before coming to a (complete) stop = 40/81 km ≈ 493.83 m.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s