the speed of a train is reduced uniformly from 10 m/s to 5 m/s while covering a distance of 150 m. What is the acceleration of the train? how much farther will it travel before coming to rest, assuming the same acceleration ?
Vo = 10 m/s.
V = 5 m/s.
d = 150 m.
a = ?
a. V^2 = Vo^2 + 2a*d
a = (V^2-Vo^2)/2d =
b. d = (V^2-Vo^2)/2a. V = 0, Vo = 5 m/s.
Well, well, well, we have a speedy train here! Let's calculate its acceleration, shall we?
To find the acceleration, we can use the formula: acceleration = change in velocity / time
Given that the speed of the train reduced uniformly from 10 m/s to 5 m/s, we can say the change in velocity is 10 m/s - 5 m/s = 5 m/s.
Now, we need to determine the time it took for this change to happen. Since we don't have that information, we'll have to find another way to calculate it.
By using the equation: distance = (initial velocity + final velocity) * time / 2
We know the initial velocity is 10 m/s, the final velocity is 5 m/s, and the distance covered is 150 m. Plugging these values into the formula, we can solve for time.
150 m = (10 m/s + 5 m/s) * time / 2
Now, let's do some quick math to find the value of time.
300 m = 15 m/s * time
time = 20 seconds
Alrighty then! Now that we know it took 20 seconds for the train to slow down, we can calculate the acceleration.
acceleration = change in velocity / time
acceleration = 5 m/s / 20 s
acceleration = 0.25 m/s²
So, the acceleration of the train is 0.25 m/s². It's like going from "I'm late" to "Why not stop and smell the flowers?"
Now, let's move on to the next part of your question. If the train keeps the same acceleration, how much farther will it travel before coming to a complete stop?
To figure that out, we'll use another formula: distance = (initial velocity^2 - final velocity^2) / (2 * acceleration)
Given that the train is coming to a stop, the final velocity is 0 m/s.
So, plugging in our values, we get:
distance = (10 m/s)^2 - (0 m/s)^2 / (2 * 0.25 m/s²)
distance = 100 m² / (0.5 m/s²)
distance = 200 meters
Therefore, the train will travel another 200 meters before it comes to a rest. It's like when you think you've reached your destination, but then you realize you forgot your wallet at home. Keep on chugging, train!
To find the acceleration of the train, we can use the formula:
Acceleration (a) = (Final Velocity (v) - Initial Velocity (u)) / Time (t)
In this case, the initial velocity (u) is 10 m/s and the final velocity (v) is 5 m/s. The time (t) is not given, but we can calculate it using the distance and formula:
Distance (d) = (Initial Velocity (u) + Final Velocity (v)) / 2 * Time (t)
Given that the distance (d) is 150 m, let's calculate the time:
150 m = (10 m/s + 5 m/s) / 2 * t
150 m = 15 m/s / 2 * t
150 m = 7.5 m/s * t
t = 150 m / 7.5 m/s
t = 20 seconds
Now we can calculate the acceleration:
Acceleration (a) = (5 m/s - 10 m/s) / 20 s
Acceleration (a) = -5 m/s / 20 s
Acceleration (a) = -0.25 m/s^2
So, the acceleration of the train is -0.25 m/s^2.
To calculate how much farther the train will travel before coming to rest, assuming the same acceleration, we can use the formula:
Distance (d) = (Final Velocity (v)^2 - Initial Velocity (u)^2) / (2 * Acceleration (a))
Given that the final velocity (v) is 0 m/s and the acceleration (a) is -0.25 m/s^2:
Distance (d) = (0 m/s^2 - 10 m/s^2)^2 / (2 * -0.25 m/s^2)
Distance (d) = (-10 m/s^2)^2 / (-0.5 m/s^2)
Distance (d) = 100 m^2 / 0.5 m/s^2
Distance (d) = 200 m
So, the train will travel an additional 200 meters before coming to rest assuming the same acceleration.
To calculate the acceleration, you can use the formula:
acceleration = (final velocity - initial velocity) / time
In this case, the initial velocity (u) is 10 m/s, the final velocity (v) is 5 m/s, and the time (t) will be determined later. So, let's proceed to find the time first.
To find the time (t), you can use the formula:
distance = (initial velocity + final velocity) / 2 * time
Rearranging the formula to solve for time:
time = distance / ((initial velocity + final velocity) / 2)
In this case, the distance (d) is 150 m. Plugging in the values:
time = 150 m / ((10 m/s + 5 m/s) / 2) = 150 m / (15 m/s / 2) = 150 m / 7.5 m/s = 20 seconds
Now that we have the time, we can calculate the acceleration:
acceleration = (final velocity - initial velocity) / time
acceleration = (5 m/s - 10 m/s) / 20 s = -5 m/s / 20 s = -0.25 m/s^2
The negative sign indicates that the train's speed is decreasing, as expected when the train is coming to a stop. Therefore, the acceleration of the train is -0.25 m/s^2.
To calculate how much farther the train will travel before coming to rest, assuming the same acceleration, we can use the following formula:
distance = (initial velocity^2 - final velocity^2) / (2 * acceleration)
In this case, the initial velocity (u) is 5 m/s, the final velocity (v) is 0 m/s, and the acceleration (a) is -0.25 m/s^2. Plugging in the values:
distance = (5 m/s)^2 - (0 m/s)^2 / (2 * -0.25 m/s^2)
distance = 25 m^2 / (-0.5 m/s^2)
distance = -50 m^2/s^2
Since the distance can't be negative, we'll take the absolute value:
distance = |-50 m^2/s^2| = 50 m
Therefore, the train will travel an additional 50 meters before coming to rest, assuming the same acceleration.