CH2. 1D Motion_ problems. 2.1 Position & Displacement: x(t) (K.CH2.15): The position of a particle as a function of time is described by the equation.

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Presentation transcript:

CH2. 1D Motion_ problems

2.1 Position & Displacement: x(t) (K.CH2.15): The position of a particle as a function of time is described by the equation x(t) = t - t 3, Where x is in m and t is in s. a)Where is the particle at t=5.0s? b)What is velocity at t = 3.0 s? c)Is acceleration constant? d) What is acceleration at t = 1.0 s? e)What is the average velocity between t=0 s and t=2.0s? f)Compute ( v(2) + v(0) )/2. is it the same as d? g)What is the maximum position position of the particle on the x axis? (answer 4.0 )

2.1 Position & Displacement: graph x(t) (K.CH2.4): Figure 2 shows the position–time graph f an object. a)Where is the particle at t=3.0s? (Ans: +20m) b)What is the average velocity of the object between t=0.0 and t=5.0? (Ans: 2m/s) c)What is the velocity of the object at t=2.0? (Ans: 6.7) d)What is the acceleration of the object at t=2.0? (Ans: 0) e)What is the total distance travelled between t = 0 and t=5.0? Average speed? (Ans:30,6m/s)

(K.CH2.4): A particle starts from the origin at t=0 and moves along the positive x-axis. A graph of the velocity of the particle as a function of time is shown in the figure. a)What is the acceleration in the time period t=0.0 till t=2.0s? (Ans: +2m/s 2 ) b)What is the average velocity of the object between t=0.0 and t=2.0? (Ans: 2.0m/s) c)What is the acceleration in the time period t=2.0 till t=5.0s? (Ans: -2m/s 2 ) d)What is the average velocity of the object between t=2.0 and t=5.0? (Ans:1.0m/s) e)What is the average velocity of the object between t=0.0 and t=5.0? (Ans: 1.4m/s)

a) Answer: 20m/s, 5m/s b) Answer: 262m

An object is thrown vertically upward with an initial speed of 98m/s. Neglect friction: a)When the maximum height is reached? b)What is the maximum height? c)What is the velocity of the object before it touch the ground? d)What is the total time of the trip? Answers a: 10 S, b: 490m, c: -98m/s, d: 20s

T max : time to reach max height2.55 S Double the t then subtract 1s4.10 S Substitute in y=y o + V o t- ½ g t 2 20m Alternative Version: What is the time to reach 10 meters before hitting the ground Use position equation with y =10: y=y o + V o t- ½ g t 2 It is a quadratic equation with two solutions Substitute in y=y o + V o t- ½ g t 2