NJC 03ES Motion and Forces Exercise Solutions 2025

National Junior College Science Department | Physics 3. Motion and Forces Exercises Solution E1 (a) displacement (direction: SW; magnitude: 200 km) (b) speed (c) velocity (direction: along straight edge of table; magnitude: 2 mm s–1) Note: The description of the direction may be unclear (not sure exactly which direction along the straight edge ) but there is undoubtedly a mention of the direction. (d) distance E2 Speedometer shows the speed your car is moving at. It does not provide information on the direction of motion. E3 Draw a vector diagram (a) distance = 3.0 + 4.0 = 7.0 km (b) magnitude of displacement |𝑠| = √32 + 42 = 5.0 km tan 𝜃 = 4 3 ⇒ 𝜃 = 53° displacement is 5.0 km 53o east of north E4 average acceleration 𝑎̅ = Δ𝑣 Δ𝑡 = 𝑣𝑓−𝑣𝑖 𝑡 = 9−0 1.5 = 6.0 m s−2 E5 average acceleration 𝑎̅ = Δ𝑣 Δ𝑡 = 𝑣𝑓−𝑣𝑖 𝑡 = 0−115×103 60×60 1.5×60 = −0.35 m s−2

National Junior College Science Department | Physics E6 speed = distance travelled time taken = 2𝜋×1.5×1011 365×24×60×60 = 3.0 × 104 m s−1 = 30 km s−1 In the course of one year, its displacement is zero, so its average velocity is zero. Note: As the Earth orbits the Sun, its direction of motion keeps changing. Hence its instantaneous velocity keeps changing. E7 average speed = distance travelled time taken = 20×2+40×2+60×6 2+2+6 = 48 m s−1 E8 s-t graph is a straight line through the origin. velocity = gradient of s-t graph = 340−0 4−0 = 85 m s−1 E9 (a) (b) acceleration for first 10 s = gradient of straight line from 0 to 10 s = 30−0 10−0 = 3.0 m s−2 (c) acceleration for last 15 s = 0−30 30−15 = −2.0 m s−2 (d) total distance = area under graph = 1 2 × (5 + 30) × 30 = 525 m

National Junior College Science Department | Physics E10 Draw a tangent to the curve at point P acceleration = 320−60 12−0 = 21.7 m s−2 E11 (a) We know u, a and t and we want to know v, so we use the equation 𝑣 = 𝑢 + 𝑎𝑡. Velocity 𝑣 = 0 + 2.0 × 10 = 20 m s−1 (b) We know u, a and t and we want to know s, so we use the equation 𝑠 = 𝑢𝑡 + 1 2 𝑎𝑡2. Distance 𝑠 = 0 × 10 + 1 2 × 2.0 × 102 = 100 m (c) We know u, v and a and we want to know t, so we rearrange the equation 𝑣 = 𝑢 + 𝑎𝑡. Time 𝑡 = 24−0 2.0 = 12 s E12 (a) We know u, v and t and we want to know a, so we use the equation 𝑣 = 𝑢 + 𝑎𝑡. Acceleration 𝑎 = 20−4.0 100 = 0.16 m s−2 (b) Average velocity 𝑣𝑎𝑣𝑔 = 𝑣+𝑢 2 = 20+4.0 2 = 12 m s−1 (c) [Method 1] Distance = average speed × time = 12 × 100 = 1200 m [Method 2] We know u, v and t and we want to know a, so we use the equation 𝑠 = 1 2 (𝑢 + 𝑣)𝑡. Distance 𝑠 = 1 2 (4.0 + 20) × 100 = 1200 m (12,320)× (0,60)×

National Junior College Science Department | Physics E13 (a) We know s = 0.80 m (↓) and a = 9.81 m s–2 (↓), and that u = 0, and we