- #1
realitybugll
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So by manipulating two of the kinematic equations you can find this relationship:
a= acceleration
t = time
v = velocity
v0 = initial velocity
x = displacement
eq (1) = v = v0 + at --> at = v - v0
eq (2) x = 1/2*(v - v0)*t --> 2x/t = v - v0
substituting 2x/t for v - v0 : at = 2x/t --> x = (at^2)/2
however when using a different method to derive this relationship i got a different answer...
Acceleration in classical mechanics is usually represented as m/s/s (meters over seconds squared) and is always constant. Acceleration can also be thought of as the rate of change of velocity, which is how we will think of it from now on.
suppose after 1 second the velocity changes by a. The velocity at 1 second must be a because a-0 = a
because acceleration is constant at 2 seconds the velocity must also change by a
and therefore the velocity at 2 seconds must be 2a because 2a - a = a
Similarly the velocity at 3 seconds must be 3a because 3a-2a = a, and the velocity at n seconds is therefore n*a
to find the total displacement we must add the velocities at each second.
for example the total displacement after 3 seconds = the velocities at 3, 2 and 1 seconds added together = 3a+2a+a = 6a
to find the equation for this we can find the first few terms of this sequence
(total displacement (x)) 1a 3a 6a 10a 15a
(seconds (t)) 1 2 3 4 5 ...
This is actually a quadratic sequence and we can find the equation for the sequence using the formula for the partial sum of a quadratic sequence. The answer we get is:
x = ((t^2 + t)*a)/2
this is different from x = (at^2)/2 and I'm wondering why...which one is correct?
The second way was probably hard to follow sorry...
a= acceleration
t = time
v = velocity
v0 = initial velocity
x = displacement
eq (1) = v = v0 + at --> at = v - v0
eq (2) x = 1/2*(v - v0)*t --> 2x/t = v - v0
substituting 2x/t for v - v0 : at = 2x/t --> x = (at^2)/2
however when using a different method to derive this relationship i got a different answer...
Acceleration in classical mechanics is usually represented as m/s/s (meters over seconds squared) and is always constant. Acceleration can also be thought of as the rate of change of velocity, which is how we will think of it from now on.
suppose after 1 second the velocity changes by a. The velocity at 1 second must be a because a-0 = a
because acceleration is constant at 2 seconds the velocity must also change by a
and therefore the velocity at 2 seconds must be 2a because 2a - a = a
Similarly the velocity at 3 seconds must be 3a because 3a-2a = a, and the velocity at n seconds is therefore n*a
to find the total displacement we must add the velocities at each second.
for example the total displacement after 3 seconds = the velocities at 3, 2 and 1 seconds added together = 3a+2a+a = 6a
to find the equation for this we can find the first few terms of this sequence
(total displacement (x)) 1a 3a 6a 10a 15a
(seconds (t)) 1 2 3 4 5 ...
This is actually a quadratic sequence and we can find the equation for the sequence using the formula for the partial sum of a quadratic sequence. The answer we get is:
x = ((t^2 + t)*a)/2
this is different from x = (at^2)/2 and I'm wondering why...which one is correct?
The second way was probably hard to follow sorry...