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.
  1. 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.
  2. 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.
  3. 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.
  4. 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.
  5. 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.
  6. 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.
  7. 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
  8. 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
  9. 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
  10. 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.
  11. 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
  12. 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

  1. B
  2. D
  3. E
  4. A
  5. C
  6. E
  7. B
  8. A
  9. C
  10. E
  11. B
  12. D