Graph of function of acceleration and velocity after time t

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The discussion revolves around graphing the velocity function influenced by a constant acceleration of 1 m/s². The user attempts to calculate velocity at various time points using the equation V final = at + V initial, but encounters discrepancies due to the need for previous outputs. They create a function table showing the velocity changes between time intervals, noting a pattern of increasing changes. The conversation highlights the importance of understanding that initial conditions can vary, and emphasizes the assumption of constant acceleration in their calculations. The user acknowledges a mistake in their approach and expresses gratitude for the assistance received.
Nano-Passion
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This is out of mere interest, though I'm having trouble figuring it out. Note, I set up this problem myself.

Homework Statement


Say I want to graph the function of velocity v at any given time t influenced by an acceleration a.

Given
a = 1 m/s^2

Find
velocity of any given time and represent it in a graph

Homework Equations



?

The Attempt at a Solution



The way I tried to solve this was by putting time t as the input (x) and velocity v as the output f(x).

I used the equation ( V final = at + v intial ) to find the velocity at any given point in time. But I found a big discrepancy because I need the previous output for the next output.

My table of functions looked something like this:

x | f(x)
--------
1 | 1
2 | 3
3 | 6
4 | 10
10| ?
 
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If v(0) = 0 (Vinitial), then for each t, try using v(t) = at + 0.
 
Notice that between t= 1 and t= 2, velocity has changed from 1 to 3, a change of 2. Between t= 2 and t= 3, velocity has changed from 3 to 6, a change of 3. Between t= 3 and t= 4, velocity has changed from 6 to 10, a change of 4. See the pattern? How much do you think the velocuity will change between t= 4 and t= 5? Add that to 10 to find the velocity at t= 5.

(This is characteristic of a linearly changing acceleration.)

And, by the way, it not necessary that "initial" mean t= 0. Here you could take the "initial postion" to be at t= 1. Then lewando's formula could be written v(t)= a(t-1)+ v(1). Of course lewando's formula assumes a constant acceleration which NOT the case here.
 
Last edited by a moderator:
Nano-Passion said:
Given
a = 1 m/s^2

This is why I thought a was constant. I thought that the function table was something N-P had generated, not part of the original problem statement.
 
lewando said:
If v(0) = 0 (Vinitial), then for each t, try using v(t) = at + 0.

Oh yes, such a silly mistake of me.

Thank you! And sorry for the late reply.
 

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