When does average velocity equal instantaneous velocity?

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SUMMARY

The discussion focuses on determining when average velocity equals instantaneous velocity using a set of time and position values. The average velocity during the first five seconds is calculated as 16/5 m/s, while the average velocities for specific intervals are derived as 2 m/s, 4 m/s, -2 m/s, and -1 m/s. The instantaneous velocity is found by taking the derivative of the position function, X(t) = At² + Bt + C, and evaluating it at specific time points. The participant confirms that constant velocity intervals yield equal average and instantaneous velocities.

PREREQUISITES
  • Understanding of kinematic equations, specifically X = At² + Bt + C and V = At + B.
  • Knowledge of calculus, particularly derivatives for finding instantaneous velocity.
  • Familiarity with average velocity calculations over time intervals.
  • Ability to interpret position vs. time graphs for slope determination.
NEXT STEPS
  • Study the concept of derivatives in calculus to enhance understanding of instantaneous velocity.
  • Learn how to graph position vs. time and analyze slopes for velocity determination.
  • Explore the relationship between average and instantaneous velocity in different motion scenarios.
  • Investigate the implications of constant velocity on average and instantaneous velocity equality.
USEFUL FOR

Students studying physics, particularly those focusing on kinematics, as well as educators and tutors seeking to clarify concepts of average and instantaneous velocity.

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instantaneous v = average v (solved)

Homework Statement



Indicate moments of time when the average velocity is equal to instantaneous velocity.

Given values

t (s) 0 | 2 | 5 | 7 | 8
x (m) 0 | 4 | 16 | 12 | 11


t (s) 0 | 2 | 5 | 7 | 8
v (m/s) 0 | 4 | 4 | 0 | -2

Homework Equations



X = At^2 + Bt + C
V = At + B

V(int) = lim delta x / delta t as t approaches zero


The Attempt at a Solution



There are two problems before this.

[a] Find average velocity of the object during first 5 second. 16/5 m/s
Find the average acceleration of the object during the same period. 4-0/5 = 4/5

I also find out the average velocity for each noticeable time intervals.

0 - A : 4-0/2-0 = 2 m/s
A - B = 16-4 / 5-2 = 4m/s
B - C = 12-16/7-5 = -2m/s
C- D = 11-12/8-7 = -1m/s

One way to find the instantaneous velocity is to take the slope of the tangent line on position vs time graph.
Another way is to find out the function of each interval and take ds/dt

So, in general, X(t) = At^2 + Bt + C

If we take V = At + B
let t = 0, and V is also 0,

V = At, and when t = 1, V = 2, 2 = A(1), A = 2, so for velocity of 0 - A time interval, I see the function is
V = 2t

This is true for 0 - A, B - C and C - D interval (i also think C - D is just part of B - C)

For A - B, it is constant velocity, so it is just a constant, V = 4

Now, coming back to solve [c] find instantaneous, I see X = At^2 + Bt + C

I use similar technique, and try to find a function for each time interval. If I do, I take dx / dt, and plug in the time, to find the instanteous velocity at any time.


I also know that constant velocity at any time = its instantaneous, but what about the rest.

So far, is my approach correct? I feel like I am doing too much here.
 
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