Free AP Physics B Practice Questions: Kinematics

AP Physics B multiple-choice questions on kinematics, with answers and detailed solutions. These questions are similar to those on the AP Physics B exam and cover concepts such as speed, distance, acceleration, and displacement. Use \(g = 10 \text{ m/s}^2\) in calculations.

1. An object is thrown vertically upward at time \(t = 0\), with an initial speed of \(40 \text{ m/s}\), from the top of an \(80 \text{ m}\) tall building. For what time interval is the object's height greater than the building's height?
   A) From \(t = 4 \text{ s}\) to \(t = 8 \text{ s}\)
   B) From \(t = 0\) to \(t = 4 \text{ s}\)
   C) From \(t = 2 \text{ s}\) to \(t = 8 \text{ s}\)
   D) From \(t = 0\) to \(t = 8 \text{ s}\)
   E) From \(t = 0\) to \(t = 6 \text{ s}\)
2. An object is thrown vertically upward at time \(t = 0\), with an initial speed of \(40 \text{ m/s}\), from the top of an \(80 \text{ m}\) tall building. At what time does the object reach its maximum height?
   A) at \(t = 8 \text{ s}\)
   B) at \(t = 4 \text{ s}\)
   C) at \(t = 2 \text{ s}\)
   D) at \(t = 1 \text{ s}\)
   E) at \(t = 0\)
3. An object is thrown vertically upward at time \(t = 0\), with an initial speed of \(40 \text{ m/s}\), from the top of a \(100 \text{ m}\) tall building. At what time does the object hit the ground?
   A) at \(t = 4 \text{ s}\)
   B) at \(t = 5 \text{ s}\)
   C) at \(t = 8 \text{ s}\)
   D) at \(t = 10 \text{ s}\)
   E) at \(t = 20 \text{ s}\)
4. A person runs \(2.5\) laps around a circular track of radius \(100 \text{ m}\) in \(2 \text{ minutes}\) and \(80 \text{ seconds}\). What is the magnitude of the average velocity?
   A) \(2.5\pi \text{ m/s}\)
   B) \(0.5 \text{ m/s}\)
   C) \(1 \text{ m/s}\)
   D) \(2 \text{ m/s}\)
   E) \(1.25\pi \text{ m/s}\)
5. A car accelerates uniformly from \(10 \text{ m/s}\) to \(30 \text{ m/s}\) in \(5 \text{ seconds}\) along a straight road. What is the car's speed at \(t = 3 \text{ s}\)?
   A) \(22 \text{ m/s}\)
   B) \(20 \text{ m/s}\)
   C) \(40 \text{ m/s}\)
   D) \(50 \text{ m/s}\)
   E) \(80 \text{ m/s}\)
6. A car accelerates uniformly from rest to \(30 \text{ m/s}\) in \(10 \text{ s}\), then moves at constant speed for \(1 \text{ minute}\), and finally decelerates uniformly to a stop in \(6 \text{ s}\). What is the total displacement?
   A) \(2,280 \text{ m}\)
   B) \(510 \text{ m}\)
   C) \(300 \text{ m}\)
   D) \(480 \text{ m}\)
   E) \(2,040 \text{ m}\)

Solutions

1. Answer: D
   The height equation is \(y(t) = 80 + 40t - 5t^2\). We solve \(y(t) > 80\): \[40t - 5t^2 > 0 \Rightarrow t(8 - t) > 0 \Rightarrow 0 < t < 8 \text{ s}.\] Thus, the object is above the building from \(t = 0\) to \(t = 8 \text{ s}\).
2. Answer: B
   At maximum height, vertical velocity is zero. Using \(v(t) = 40 - 10t = 0\), we get \(t = 4 \text{ s}\).
3. Answer: D
   The height equation is \(y(t) = 100 + 40t - 5t^2\). Setting \(y(t) = 0\): \[-5t^2 + 40t + 100 = 0 \Rightarrow t^2 - 8t - 20 = 0.\] Solving gives \(t = 10 \text{ s}\) (positive root).
4. Answer: C
   Total time: \(2 \text{ min } 80 \text{ s} = 200 \text{ s}\). After 2.5 laps, the displacement is the diameter (\(200 \text{ m}\)). Average velocity = displacement / time = \(200 \text{ m} / 200 \text{ s} = 1 \text{ m/s}\).
5. Answer: A
   Acceleration \(a = (30 - 10)/5 = 4 \text{ m/s}^2\). Then \(v(3) = 10 + 4 \times 3 = 22 \text{ m/s}\).
6. Answer: E
   
   • Phase 1 (acceleration): distance = average velocity × time = \(\frac{0+30}{2} \times 10 = 150 \text{ m}\).
   • Phase 2 (constant speed): distance = \(30 \times 60 = 1800 \text{ m}\).
   • Phase 3 (deceleration): distance = \(\frac{30+0}{2} \times 6 = 90 \text{ m}\).
   Total displacement = \(150 + 1800 + 90 = 2040 \text{ m}\).