# Free SAT II Physics Solutions Graphical Analysis of Motion

 Solutions with with detailed explanations to Free SAT II Physics Practice Questions on Graphical Analysis of Motion. Question 1 to 4 refer to the displacement vs time below. Fig1. - Displacement versus time. The graph of the position x versus time t of a moving object is shown in figure 1 above . On which time interval(s) is the velocity of the moving object equal to zero? A) (0 , 1) B) (1 , 4) C) (4 , 6) D) (6 , 9) E) (9 , 11) Solution - Explanations The position of moving object does not change between t = 1 s and t = 4 s and therefore the velocity is equal to zero in this interval of time. Answer B. The graph of the position x versus time t of a moving object is shown in figure 1 above . At what time was the object furthest from the origin (x = 0)? A) t = 1 s B) t = 4 s C) t = 6 s D) t = 9 s E) t = 11 s Solution - Explanations At t = 9 s, x = -15 m and this is the furthest point from x = 0. Answer D. The graph of the position x versus time t of a moving object is shown in figure 1 above . Over which time interval(s) was the object moving in the negative direction? A) (6 , 11) B) (0 , 1) C) (1 , 4) D) (4 , 11) E) (4 , 9) Solution - Explanations From t = 4 s to t = 9 s, x is decreasing and therefore the velocity is negative which means the object is moving in the negative direction. Answer E. The graph of the position x versus time t of a moving object is shown in figure 1 above . Over which time interval(s) was the object moving in the positive direction? A) (0 , 1) and (9 , 11) B) (0 , 6) C) (1 , 4) and (6 , 11) D) (0 , 4) and (4 , 9) E) (0 , 11) Solution - Explanations In both intervals 0 to 1 and 9 to 11, x is increasing, the velocity is positive and therefore the object is moving in the positive direction. Answer A. Question 5 to 10 refer to the displacement vs time below. Fig2. - Velocity versus time. The graph of the velocity v versus time t of a moving object is shown in figure 2 above. What is the total displacement from t = 0 to t = 9 seconds? A) -2.5 m B) 2.5 m C) 0 m D) 9 m E) 15 m Solution - Explanations The displacement is given by the area between the t-axis and the graph of the velocity. Area of Trapezoid on the left below t-axis = -(1/2)(4 + 2)(2.5) = -7.5 Area of Triangle on the right above t-axis = (1/2)(3)(5) = 7.5 total displacement from t= 0 to t = 9 = Total area = -7.5 + 7.5 = 0 Answer C. The graph of the velocity v versus time t of a moving object is shown in figure 2 above. What is the total distance covered by the object from t = 0 to t = 9 seconds? A) 9 m B) 5 m C) 7.5 m D) 30 m E) 15 m Solution - Explanations Between t = 0 and t = 4, the object is moving in the negative direction (velocity negative). The distance is given by the absolute value of the displacement which is given by the area. distance from (t=0 to t=4) = | -(1/2)(4 + 2)(2.5) | = 7.5 Between t = 6 and t = 9, the object is moving in the positive direction (velocity positive). The distance is given by the absolute value of the displacement which is given by the area. distance from (t = 6 to t = 9) = | (1/2)(3)(5) | = 7.5 total distance from (t= 0 to t = 9) = 7.5 + 7.5 = 15 m Answer E. The graph of the velocity v versus time t of a moving object is shown in figure 2 above. What is the average velocity over the interval t = 0 to t = 9 seconds? A) 0.6 m/s B) 0 m/s C) 5 m/s D) 1.5 m/s E) 9 m/s Solution - Explanations Note: the displacement was found to be zero in question 5 above. average velocity = displacement / time = 0 / 9 = 0 Answer B The graph of the velocity v versus time t of a moving object is shown in figure 2 above. What is the average speed over the interval t = 0 to t = 9 seconds? A) 1.7 m/s B) 9 m/s C) 5 m/s D) 7.5 m/s E) 15 m/s Solution - Explanations Note: the distance was found to be 15 m in question 6 above. average speed = distance / time = 15 / 9 = 1.7 m/s (rounded to 2 sf) Answer A The graph of the velocity v versus time t of a moving object is shown in figure 2 above. Over which time interval(s) was the object accelerating uniformly? A) (6 , 9) B) (4 , 6) C) (3 , 4) and (6 , 7) D) (0 , 1) and (7 , 9) E) (0 , 9) Solution - Explanations Uniform acceleration happens when the velocity increases linearly with time. According to the graph above, uniform acceleration in the intervals (3 , 4) and (6 , 7). Answer C The graph of the velocity v versus time t of a moving object is shown in figure 2 above. Over which time interval(s) was the acceleration of the object equal to zero? A) (1 , 4) B) (4 , 7) C) (1 , 4) D) (6 , 9) E) (1 , 3) and (4 ,6) Solution - Explanations Acceleration is equal to zero if the velocity is constant. According to the graph the acceleration is zero in the intervals (1 , 3) and (4 ,6). Answer E Question 11 to 12 refer to the acceleration vs time below. Fig3. - Acceleration versus time. The graph of the acceleration v versus time t of a moving object is shown in figure 3 above. Over which time interval(s) was the object accelerating uniformly? A) (0 , 9) B) (0 , 2) C) (2 , 4) D) (2 , 9) E) (4 , 8) Solution - Explanations Uniform acceleration, constant and positive, is in the interval (0 , 2). Answer B The graph of the acceleration v versus time t of a moving object is shown in figure 3 above. Over which time interval(s) was the object decelerating uniformly? A) (0 , 2) B) (2 , 4) C) (8 , 9) D) (4 , 8) E) (2 , 9) Solution - Explanations Uniform deceleration, constant and negative, is in the interval (4 , 8). Answer D Answers to the Above questions B D E A C E B A C E B D