![S(t) Displacement The location of the particle at a point in time [ m., ft. , cm.]S(t) > 0, the particle is to the right of 0 [above]S(t) < 0, the particle is to the left of 0 [below]S(t) = 0 at 0 (Initial displacement)](https://image.slidesharecdn.com/particlemovement-110518112132-phpapp01/85/Particle-movement-2-320.jpg)
![V(t) Velocity V(t) = s’(t)The direction the particle is moving [m/sec, ft./min]V(t) >0, the particle is moving right [up]V(t) <0, the particle is moving left [down]V(t)= 0, the particle is stopped [not moving]](https://image.slidesharecdn.com/particlemovement-110518112132-phpapp01/85/Particle-movement-3-320.jpg)
Particle movement
- 1. Particle Movement By Jillian Calhoun & Sherri Schadler
- 2. S(t) Displacement The location of the particle at a point in time [ m., ft. , cm.]S(t) > 0, the particle is to the right of 0 [above]S(t) < 0, the particle is to the left of 0 [below]S(t) = 0 at 0 (Initial displacement)
- 3. V(t) Velocity V(t) = s’(t)The direction the particle is moving [m/sec, ft./min]V(t) >0, the particle is moving right [up]V(t) <0, the particle is moving left [down]V(t)= 0, the particle is stopped [not moving]
- 4. A(t) Acceleration A(t) = V’(t)= S “(t)A(t) >0, the velocity of the particle is increasingA(t) <0, the velocity of the particle is decreasingA(t) =0, the velocity of the particle may be at a max/min.
- 5. Example If given the problem s(t) = t²- 4t + 3 t ≥ 0 seconds then, To solve you would first find the initial displacement…(set t= 0) S(0)= 0² - 4(0) + 3 = cm The initial displacement is equal to 3cm.
- 6. V(t) = s’(t) = 2t- 4 and t will = 2 When 2 is placed on a number line it shows that it is decreasing from 0-2 and increasing from 2 on . This means that the particle changes direction at t=2. Then you would plug 2 in the displacement function to determine where the change occurs. S(2) = 2²- 4(2) + 3 s=-1
- 7. A(t) = v’(t) = s”(t) The second derivative of s(t) will equal 2. a(t)=20 ≠ 2 So velocity will be increasing CONCLUSION:S(0) = 3cm The particle is starting 3cm to the right of 0. V(0) = -4cm/sec The particle then moves left at 4 cm/sec and then changes directions. Velocity is increasing and a(0) = 2cm²/sec. This shows the speed of the particle at 0.
- 8. Working Backwards If you are not given the initial displacement first; you must then take the anti derivative or integrand of either a(t) or v(t). So if you were first given v(t)= 2t-4 you must do the anti-derivative in order to find s(t). S(0)= 3cm. S(t) =∫ 2t-4 2t²/2 – 4t +C t²-4t+C3=0² - 4(0)+C C= 3 S(t) = t²-4t +3 C= constant
- 9. Notes When Velocity is increasing the acceleration is positive. Speed is decreasing when acceleration and velocity have different signs When graphing, a maximum height occurs when v= 0 Particle movement can also be related to dropping a ball. You then need to take in consideration that gravity is -9.8 m/sec² and -32ft./sec².