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.

displacement vs time
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.
    velocity vs time
    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.
    acceleration vs time
    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

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