Kinematics: Unit Diagnostic (Physics 1 Unit 1)

Answer: (A). Add the arrows tip to tail as signed numbers: (+4) + (−7) + (+1) = −2 squares. With 2.0 m per square, that is −4.0 m: 4.0 m to the left.

• (B) gets the size right but loses the sign. The long leftward move outweighs the two rightward moves.
• (C) adds the lengths of all three arrows. That is the distance travelled (24 m), a scalar, and it has no direction.
• (D) reads the answer in squares and forgets to apply the scale.

If you missed this: read “Adding vectors in one dimension” in the Topic 1.1 study guide.

Answer: (A). Average acceleration is Δv_x ÷ Δt. The four values are (+3.0) ÷ 1.0 = +3.0 m/s², (−8.0) ÷ 4.0 = −2.0 m/s², 0 ÷ 1.0 = 0 and (−4.0) ÷ 2.0 = −2.0 m/s². The largest size is 3.0 m/s².

• (B) has the largest change in velocity (8.0 m/s), and the scooter reverses. But that change is spread over 4.0 s, so the rate is smaller.
• (C) is where the scooter moves backward, but its velocity does not change (−3.0 m/s throughout), so the acceleration is zero.
• (D) is where the speed reaches its largest value, 7.0 m/s. A large speed is not a large acceleration.

If you missed this: read “Average acceleration” in the Topic 1.2 study guide.

Answer: (A). The slope is (−2.0 − 6.0) ÷ 4.0 = −2.0 m/s², so v_x reaches zero at t = 3.0 s. Up to then the cart moves right. The area of that triangle is ½ × 3.0 × 6.0 = +9.0 m. After 3.0 s the cart moves back left (area −1.0 m), so it never gets farther than 9.0 m from its start.

• (B) is the net displacement at the end, 9.0 − 1.0 = 8.0 m. The cart was farther away at 3.0 s.
• (C) is the total distance travelled, 9.0 + 1.0 m. Distance is not separation from the start.
• (D) treats the area below the axis as positive and uses the whole trapezium, ½(6.0 + 2.0) × 4.0.

If you missed this: read “Graphs: slopes and areas” in the Topic 1.3 study guide. The cart turns round where the line crosses the time axis.

Answer: (C). From rest, x = ½a_x t². So 2.0 = ½a_x(1.0)², which gives a_x = 4.0 m/s². At 2.0 s, x = ½(4.0)(2.0)² = 8.0 m. During the second second it travels 8.0 − 2.0 = 6.0 m.

• (A) assumes constant velocity, but the sled is speeding up.
• (B) assumes distance grows in proportion to time. With constant acceleration from rest, it grows with t².
• (D) is the total distance after 2.0 s, not the distance in the second interval alone.

If you missed this: read “Where the constant-acceleration equations come from” in the Topic 1.3 study guide.

Answer: (A). At 5.0 s the bus is at 40 + 12 × 5.0 = 100 m and the cyclist is at 8.0 × 5.0 = 40 m, both measured from the sign. So x_bus,cyclist = 100 − 40 = +60 m: 60 m ahead. In the cyclist’s frame the bus gains 4.0 m/s, and 40 + 4.0 × 5.0 = 60 m agrees.

• (B) is the bus’s position measured from the sign. It ignores the observer’s own motion.
• (C) adds the two velocities, which is only right when the bus and cyclist move in opposite directions.
• (D) is the cyclist’s position measured by the bus. The subtraction is the wrong way round.

If you missed this: read “Converting positions and displacements” in the Topic 1.4 study guide.

Answer: (B). Treat each axis on its own. v_x stays 6.0 m/s. v_y = 8.0 − 9.8 × 1.0 = −1.8 m/s, so the ball is already falling. Speed = √(6.0² + 1.8²) = 6.3 m/s.

• (A) is the launch speed, √(6.0² + 8.0²). It assumes the speed never changes.
• (C) is the speed at the top, where v_y = 0. The top came earlier, at 8.0 ÷ 9.8 ≈ 0.82 s.
• (D) uses only the vertical component and forgets the horizontal motion.

If you missed this: read “Projectile motion” and Worked example 3 in the Topic 1.5 study guide.

Answer: (A). Size: √(4.0² + 3.0²) = 5.0 m/s. Both components are negative, so a sketch puts the vector in the south-west quadrant. The angle below the west direction is tan⁻¹(3.0 ÷ 4.0) = 37°.

• (B) trusts the calculator: tan⁻¹((−3.0) ÷ (−4.0)) also gives 37°, but the two minus signs cancel and hide the quadrant. Always sketch.
• (C) adds the sizes of the components. Perpendicular components combine by Pythagoras.
• (D) uses tan⁻¹(4.0 ÷ 3.0), which measures the angle from the south direction. “53° west of south” would be right; “53° south of west” is not.

If you missed this: read “Rebuilding a vector from its components” in the Topic 1.5 study guide.

(a) v_x–t: a straight line from 0 to 9.0 m/s at 2.5 s, then a horizontal line at 9.0 m/s. a_x–t: flat at 9.0 ÷ 2.5 = 3.6 m/s² from 0 to 2.5 s, then zero.

(b) Stage 1 distance = triangle area = ½ × 2.5 × 9.0 = 11.25 m. The remaining 60 − 11.25 = 48.75 m takes 48.75 ÷ 9.0 = 5.42 s. Total 7.9 s.

(c) The equation needs constant acceleration over the whole interval. Here a_x = 3.6 m/s² only for the first 2.5 s and zero afterwards. Using it for the whole run assumes the sprinter keeps speeding up, which would mean reaching about 21 m/s.

Check yourself: 1 point for each graph with values, 1 for splitting the run into two stages, 1 for 7.9 s, 1 for the constant-acceleration condition in (c).

If you missed this: work through Figure 2 and the equations table in the Topic 1.3 study guide.

(a) v_x0 = 14 cos 35° = 11.47 m/s, and v_x never changes. t = 10 ÷ 11.47 = 0.87 s.

(b) v_y0 = 14 sin 35° = 8.03 m/s. y = v_y0 t − ½gt² = 8.03(0.872) − 4.9(0.872)² = 7.00 − 3.73 = 3.3 m.

(c) v_y = 8.03 − 9.8(0.872) = −0.52 m/s. The sign is negative, so the ball is just falling: it passed its highest point at 8.03 ÷ 9.8 = 0.82 s, a moment before reaching the wall.

Check yourself: 1 point for using only the horizontal motion to find t, 1 for 3.3 m, 1 for the sign of v_y as the reason in (c). A common error is y = v_y0 t = 7.0 m, which leaves out gravity.

If you missed this: read “Motion in two dimensions: one problem becomes two” in the Topic 1.5 study guide.