First off, I believe you mean to ask the deceleration of the train. Assuming it was leaving a stop, that would be acceleration. But in this case, it's deceleration since it's arriving at a stop.
My aim is to use right terms.
From the fourth equation of motion:
v^2 = u^2 + 2as
where v is final velocity or speed — in this 0m/s (since the train comes to a stop)
u is initial velocity — in this case 4m/s
a is acceleration which is unknown
s is the distance (in this case 100m)
Making a, the subject of the formula (since this is our unknown), we have:
a = (v^2 — u^2) / 2s
Substituting our values in, we arrive at:
a = (0^2 — 4^2) / 2*100
By the time you calculate this, your final answer becomes —0.08.
Therefore, your acceleration is —0.08m/s^2. But the better way to put it (using right terms) would be that the deceleration of the train is 0.08m/s^2.
Upvote if you find this helpful.
Feel free to ask me if there's anything you don't understand from my explanation.
The following formula will allow you to calculate the acceleration of the train.
vf² = vi² + 2ad; where vf = final velocity, vi = initial velocity, a = acceleration, and d = dísplacement
vf = 0 m/s
vf² = (0 m/s)² = 0 m²/s²
vi = 4 m/s
vi² = (4 m/s)² = 16 m²/s²
a = ? m/s²
d = 100 m
Calculate acceleration.
a = (vf² - vi²)/2d
a = (0 m²/s² - 16 m²/s²)/(2 × 100 m) = -0.08 m/s²
Negative acceleration means the train slowed down.
S = 100 m
Vf =0
Vi = 4 m/s
a = (Vf^2 - Vi^2)/2*S = (0–16m^2/s^2)/200 m = -.08m/s^2
The train is decelerating at the rate of 0.08 m/s^2
Answers & Comments
Answer:
First off, I believe you mean to ask the deceleration of the train. Assuming it was leaving a stop, that would be acceleration. But in this case, it's deceleration since it's arriving at a stop.
My aim is to use right terms.
From the fourth equation of motion:
v^2 = u^2 + 2as
where v is final velocity or speed — in this 0m/s (since the train comes to a stop)
u is initial velocity — in this case 4m/s
a is acceleration which is unknown
s is the distance (in this case 100m)
Making a, the subject of the formula (since this is our unknown), we have:
a = (v^2 — u^2) / 2s
Substituting our values in, we arrive at:
a = (0^2 — 4^2) / 2*100
By the time you calculate this, your final answer becomes —0.08.
Therefore, your acceleration is —0.08m/s^2. But the better way to put it (using right terms) would be that the deceleration of the train is 0.08m/s^2.
Upvote if you find this helpful.
Feel free to ask me if there's anything you don't understand from my explanation.
The following formula will allow you to calculate the acceleration of the train.
vf² = vi² + 2ad; where vf = final velocity, vi = initial velocity, a = acceleration, and d = dísplacement
vf = 0 m/s
vf² = (0 m/s)² = 0 m²/s²
vi = 4 m/s
vi² = (4 m/s)² = 16 m²/s²
a = ? m/s²
d = 100 m
Calculate acceleration.
a = (vf² - vi²)/2d
a = (0 m²/s² - 16 m²/s²)/(2 × 100 m) = -0.08 m/s²
Negative acceleration means the train slowed down.
S = 100 m
Vf =0
Vi = 4 m/s
a = (Vf^2 - Vi^2)/2*S = (0–16m^2/s^2)/200 m = -.08m/s^2
The train is decelerating at the rate of 0.08 m/s^2
Explanation:
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