The problems are on the average speed worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Average speed is the total distance an object travels divided by the total time it takes, written v = d ÷ t. These worked answers correct the 20 problems of the grade 6 science worksheet on average speed, from moving or at rest? to laps of a track: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the final result with its unit in bold. Compare your method as well as your result, and check that you converted minutes into hours or hours into seconds wherever needed.
Level: Grade 6 science (Year 7 · Class 6) · Curriculum: MS-PS2-2
Problem 1 Moving or at rest?
1. Relative to the seat she is at rest.2. Relative to the tree she is moving.
3. Whether an object moves depends on the reference object: her position does not change compared with the seat, but it changes compared with the tree.
Problem 2 Speed formula and units
1. Average speed = distance ÷ time, so v = d ÷ t.2. The SI unit is metre per second (m/s).
3. Kilometre per hour (km/h) is common on roads (miles per hour is used in the UK and USA).
Problem 3 A 100 m run
v = d ÷ t = 100 m ÷ 20 s = 5 m/s.Problem 4 A long drive
v = 150 km ÷ 3 h = 50 km/h.Problem 5 Reading dotted tracks
1. A is a straight line, B is a circle and C is a curve (an arc).2. The tip of a second hand moves in a circle, so track B.
3. Equal distances in equal times means the speed is constant.
Problem 6 Distance of a scooter
d = v × t = 12 m/s × 30 s = 360 m.Problem 7 Time for a bus journey
1. t = d ÷ v = 200 ÷ 80 = 2.5 h.2. 0.5 h = 30 min, so the journey takes 2 h 30 min.
Problem 8 Converting units
1. 54 ÷ 3.6 = 15 m/s.2. 8 × 3.6 = 28.8 km/h.
3. 29 × 3.6 = 104.4 km/h, which is more than 100 km/h.
Problem 9 Cycling to the market
1. 20 min = 20 ÷ 60 = 1/3 h.2. v = 6 ÷ (1/3) = 18 km/h.
3. v = 6000 m ÷ 1200 s = 5 m/s (check: 18 ÷ 3.6 = 5).
Problem 10 Reading a graph
1. After 20 s the walker has covered 30 m.2. 45 m is reached after 30 s.
3. v = 60 m ÷ 40 s = 1.5 m/s.
Problem 11 Who is faster?
1. Aarav: 400 ÷ 80 = 5 m/s. Hina: 2700 m ÷ 600 s = 4.5 m/s.2. Aarav is faster because 5 m/s is greater than 4.5 m/s.
Problem 12 Journey with a stop
1. v = 90 ÷ 2 = 45 km/h.2. Moving time = 2 h − 0.5 h = 1.5 h, so v = 90 ÷ 1.5 = 60 km/h.
3. The stop counts in the total time but adds no distance, so the average over the whole journey is lower.
Problem 13 Spot the error
1. The time was left in minutes while the speed was wanted in km/h, so the result is wrong (and far too small for a car).2. 90 min = 1.5 h, so v = 150 ÷ 1.5 = 100 km/h.
Problem 14 Toy car data
1. 0-2 s: 4 ÷ 2 = 2 m/s. 2-4 s: 8 ÷ 2 = 4 m/s. 4-6 s: 4 ÷ 2 = 2 m/s. 6-8 s: 0 m/s.2. The car was at rest between 6 s and 8 s.
3. v = 16 ÷ 8 = 2 m/s.
Problem 15 A sprint in two units
1. v = 100 ÷ 9.8 = 10.204… = 10.2 m/s.2. 10.204 × 3.6 = 36.73… = 36.7 km/h.
Problem 16 Two parts of a trip
1. First part: 60 ÷ 40 = 1.5 h. Second part: 100 ÷ 80 = 1.25 h.2. Total distance 160 km, total time 2.75 h, so v = 160 ÷ 2.75 = 58.2 km/h.
3. The mean of 40 and 80 is 60 km/h, which is different, because more time was spent at the lower speed.
Problem 17 Cyclists meeting
1. The gap shrinks at 12 + 15 = 27 km/h.2. t = 27 ÷ 27 = 1 h.
3. d = 12 × 1 = 12 km (Diego has covered 15 km, and 12 + 15 = 27 km).
Problem 18 Two runners on a graph
1. A: 100 ÷ 20 = 5 m/s. B: (80 − 20) ÷ 20 = 3 m/s.2. B starts 20 m ahead.
3. The lines cross where 5t = 20 + 3t, so 2t = 20 and t = 10 s. Then d = 5 × 10 = 50 m from the start.
Problem 19 Catching the school bus
1. t = 18 ÷ 40 = 0.45 h = 0.45 × 60 = 27 min.2. 07:45 minus 27 min = 07:18.
Problem 20 Laps of a track
1. d = 3 × 400 = 1200 m.2. t = 270 s, so v = 1200 ÷ 270 = 4.44… = 4.4 m/s; 4.444 × 3.6 = 16 km/h.
3. Average speed uses the total distance travelled along the trajectory, not the straight-line gap between start and finish, so it is not zero.
