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1D Motion - Position as a Funtion of Time

  • Thread starter Wboyt92
  • Start date
All, I am reviewing for a comprehensive exam and am extremely weak on 1D Motion. Can you please help me out?

A lab cart travels along the x-axis and its position as a function of time is given by the following expression: x(t)= 6m - (8 m/s)t + (1 m/s2)t2

1. Displacement of cart during time interval t=0s to t=2s
2. [tex]\Delta[/tex]x of cart during time interval t=2s to t=3s.
3. [tex]\Delta[/tex]x of cart during time interval t=0s to t=5s.
4. Total distance from t=0s to t=5s.
5. Initial velocity (@t=0) in m/s.
6. Average velocity from t=0s to t=5s.
7. Average speed (m/s) t=0s to t=5s.
8. Instantaneous velocity (m/s) @ t=3s.
9. Acceleration in (m/s2)
10. Where (if ever) does the cart change direction?
You have:

delta x can be calculated with simple substitution.

You need to find an equation for both velocity and acceleration by means of differentiation.

Recall the definitions for the terms: average, velocity and speed (a plot might help).

Direction changes when velocity is zero.
for questions 1-5, plug in the time given to for t then use those numbers to complete the problem. to find equations for velocity and acceleration, you must differentiate the position function.

ex. x(t)=d(1+2t+4t^2) --> v(t)=8t+2 then v(t)=d(8t+2) --> a(t)=8
position function becomes velocity function _velocity becomes acceleration function

just use the equation given to you, and in the acceleration equation, a lone number indicates a constant acceleration.

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