What mathematical formula is needed to determine the student's displacement from point A to C? (Hint: think
1 answer:
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To explain, I will use the equations for kinetic and potential energy:
Potential energy is the potential an object has to move due to gravity. An object can only have potential energy if 1) <u>gravity is present</u> and 2) <u>it is above the ground at height h</u>. If gravity = 0 or height = 0, there is no potential energy. Example:
An object of 5 kg is sitting on a table 5 meters above the ground on earth (g = 9.8 m/s^2). What is the object's gravitational potential energy? <u>(answer: 5*5*9.8 = 245 J</u>)
(gravitational potential energy is potential energy)
<h3>Kinetic energy</h3>Kinetic energy is the energy of an object has while in motion. An object can only have kinetic energy if the object has a non-zero velocity (it is moving and not stationary). An example:
An object of 5 kg is moving at 5 m/s. What is the object's kinetic energy? (<u>answer: 5*5 = 25 J</u>)
<h3>Kinetic and Potential Energy</h3>Sometimes, an object can have both kinetic and potential energy. If an object is moving (kinetic energy) and is above the ground (potential), it will have both. To find the total (mechanical) energy, you can add the kinetic and potential energies together. An example:
An object of 5 kg is moving on a 5 meter table at 10 m/s. What is the objects mechanical (total) energy? (<u>answer: KE = .5(5)(10^2) = 250 J; PE = (5)(9.8)(5) = 245 J; total: 245 + 250 = 495 J</u>)
Answer:
(i) 12 seconds
(ii) 216 meters from the initial position
(iii) 132 meters from the initial position
(iv) No
Explanation:
Speed of express train =36 m/s
Speed of local train =11 m/s
The initial distance between the local train and passenger train =100 m.
Due to the application of breaks, the express train slows at the rare of
So, the acceleration of the express train, .
(i) Let t be the time the express train takes to stop.
From the equation of motion,
v=u+at
where, v: final velocity, u: initial velocity, a: constant acceleration, t: time taken to change the speed from u to v.
In this case, v=0, u=36 m/s,
So, 0=36+(-3)t
seconds.
(ii) To compute the distance traveled, s, till the express train stops, using
meters.
(iii) The local train is moving at a speed of 11 m/s
So, in 12 seconds, the distance, d, traveled by the local train
d= 11x12=132 meters [as distance= speed x time]
(iv) Let 0 be the reference position which is the initial position of the express train.
So, at the initial time, the position of the local train is at 100m.
After 12 seconds:
The position of the express train is at 216 m [using part (ii)]
and the position of the local train is at 100+132=232m [using part (iii)].
So, the local train is still ahead of the express train, hence the trains didn't collide.