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Is stepping on the brakes of a moving train an example of acceleration?
No, stepping on the brakes of a moving train is an example of deceleration, as it is the action of slowing down or reducing the speed of the train. Acceleration refers to an increase in speed or velocity.
Which describes a train and acceleration if it changes speed from 25 ms to 10 ms in 240 s?
The train is decelerating since its speed is decreasing from 25 m/s to 10 m/s. The acceleration can be calculated using the formula: acceleration = (final velocity - initial velocity) / time. Substituting the values, the acceleration would be approximately -0.06 m/s^2.
Which describes a trains acceleration if it changes speed from 25ms to 10ms in 240s?
The acceleration of the train can be calculated using the formula: acceleration = (final velocity - initial velocity) / time. Plugging in the values: (10 m/s - 25 m/s) / 240 s = -0.0625 m/s^2. The negative sign indicates that the train is decelerating.
If a train is accelerating at a rate of 4km per hour per second and its initial velocity is 20km per hour what is its velocity after 30 seconds?
The train's velocity after 30 seconds can be calculated using the formula: final velocity = initial velocity + (acceleration * time). Plugging in the values, final velocity = 20 km/hr + (4 km/hr/s * 30 s) = 20 km/hr + 120 km/hr = 140 km/hr. So, the train's velocity after 30 seconds is 140 km/hr.
What describes a train's acceleration if it changes speed from 25 ms to 10 ms in 240 s?
The train's acceleration would be -0.06 m/s^2. This is obtained by using the formula: acceleration = (final velocity - initial velocity) / time. Plugging in the values gives: (-10 - 25) / 240 = -35 / 240 = -0.06.
A train travels from one city to another its initial velocity is lower then its final velocity?
Positive acceleration.
If the speeding train hits the brakes and it takes the train 39 seconds to go from 54.8 ms to 12 ms what is the acceleration?
Well, isn't that a happy little math problem we have here! To find the acceleration, we can use the formula: acceleration = (final velocity - initial velocity) / time. Plugging in the values, we get: acceleration = (12 - 54.8) / 39. So, the acceleration of the train is approximately -1.27 m/s². Remember, there are no mistakes in math, just happy little accidents!
Is stepping on the brakes of a moving train an example of acceleration?
No, stepping on the brakes of a moving train is an example of deceleration, as it is the action of slowing down or reducing the speed of the train. Acceleration refers to an increase in speed or velocity.
What is the maximum acceleration of a train travelling at 90 miles per hour?
The speed or velocity of a train has no bearing on its acceleration.
Which describes a train and acceleration if it changes speed from 25 ms to 10 ms in 240 s?
The train is decelerating since its speed is decreasing from 25 m/s to 10 m/s. The acceleration can be calculated using the formula: acceleration = (final velocity - initial velocity) / time. Substituting the values, the acceleration would be approximately -0.06 m/s^2.
Which describes a trains acceleration if it changes speed from 25ms to 10ms in 240s?
The acceleration of the train can be calculated using the formula: acceleration = (final velocity - initial velocity) / time. Plugging in the values: (10 m/s - 25 m/s) / 240 s = -0.0625 m/s^2. The negative sign indicates that the train is decelerating.
Can a body at rest have an accleration?
acceleration is a relative quantity . state of rest or motion is also relative . if two body is in rest or moving with same velocity and having same acceleration then one is in state of rest with respect to other . suppose a person sitting in a train then he is in rest with respect to train but he is moving with the acceleration of train with respect to the ground.
If a bird hovers inside a train whilst that train is moving then presumably it is travelling at the same speed as the train Why is this and why is it not thown to the back of a carriage?
What pushes us back is a change in velocity (i.e., an acceleration), not the velocity itself. You might as well say that the train is stationary, and that planet Earth is moving under the train. In outer space, there is no fixed reference point; and it doesn't make sense to speak of the "real" velocity. A velocity must always be indicated with relationship to some reference point. Using the Earth as a reference point is convenient, but it isn't the only option.What pushes us back is a change in velocity (i.e., an acceleration), not the velocity itself. You might as well say that the train is stationary, and that planet Earth is moving under the train. In outer space, there is no fixed reference point; and it doesn't make sense to speak of the "real" velocity. A velocity must always be indicated with relationship to some reference point. Using the Earth as a reference point is convenient, but it isn't the only option.What pushes us back is a change in velocity (i.e., an acceleration), not the velocity itself. You might as well say that the train is stationary, and that planet Earth is moving under the train. In outer space, there is no fixed reference point; and it doesn't make sense to speak of the "real" velocity. A velocity must always be indicated with relationship to some reference point. Using the Earth as a reference point is convenient, but it isn't the only option.What pushes us back is a change in velocity (i.e., an acceleration), not the velocity itself. You might as well say that the train is stationary, and that planet Earth is moving under the train. In outer space, there is no fixed reference point; and it doesn't make sense to speak of the "real" velocity. A velocity must always be indicated with relationship to some reference point. Using the Earth as a reference point is convenient, but it isn't the only option.
What is the acceleration of any object moving at constant velocity?
0. Doesn't matter what unit it is. If it's moving at a constant velocity, not changing its speed (either positively or negatively), it's not accelerating, right? So its acceleration is 0. However, we must remember to always define; 'with respect to what'. Velocity is a relative concept. i.e. If you are sitting at rest or walking with constant velocity on a train, yet the train is accelerating, are you accelerating? wrt the train - the answer is no. wrt the embankment - the answer is yes. The answer then relates to something else, which is your own 'centre of mass' inertial rest frame. (i.e. you can 'feel' acceleration). So wrt your 'previous' state. This is normally quite poorly understood.
How do you make a complaint to corporate headquarters for Days Inn motels?
a train staring from rest acquires a velocity of 40m/s in 10sec.what is its acceleration
If a train is accelerating at a rate of 4km per hour per second and its initial velocity is 20km per hour what is its velocity after 30 seconds?
The train's velocity after 30 seconds can be calculated using the formula: final velocity = initial velocity + (acceleration * time). Plugging in the values, final velocity = 20 km/hr + (4 km/hr/s * 30 s) = 20 km/hr + 120 km/hr = 140 km/hr. So, the train's velocity after 30 seconds is 140 km/hr.
What describes a train's acceleration if it changes speed from 25 ms to 10 ms in 240 s?
The train's acceleration would be -0.06 m/s^2. This is obtained by using the formula: acceleration = (final velocity - initial velocity) / time. Plugging in the values gives: (-10 - 25) / 240 = -35 / 240 = -0.06.
