Σ/ε
σ = F/A
ε = ΔL/L
F = force
A = area
L = lenght
ΔL = |old lenght - new lenght|
Answer:
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.
Explanation:
It is given that,
The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation as:

Where,
x is in meters and t is in sec
We know that,
Velocity,

(a) i. t = 2 s

At t = 4 s

(b) Acceleration,

Pu t = 3 s in the above equation
So,

Hence, this is the required solution.
Answer:
600 J
Explanation:
it's obviously btw so yeahhhh
Answer is miles sir your welcome it was simple