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.

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.





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.





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