Flashcards
UNIT 6.1 DISTANCE, SPEED AND ACCELERATION
Flashcards do conjunto (10)
- 1A car travels 2 km in 80 seconds. How fast was it travelling in meters per second? (Remember 1 km = 1,000 m)S = D ÷ T
S = 2,000 ÷ 80
S = 25 m/s - 2A rocket travels at 1,500 m/s for 20 minutes. How far has it travelled in meters? (Remember 1 minute = 60 seconds)T = 20 mins = 1,200 seconds
D = S x T
D = 1,500 x 1,200
D = 1,800,000 m - 3The moon is 384,000 km away. How long would it take to get there if you travelled at 100 km/h?T = D ÷ S
T = 384,000 ÷ 100
T = 3,840 hours or 160 days - 4Interpret this distance-time graph.A to B: The object moves at a constant speed for 5 hours.
B to C: The object stops moving for 3 hours.
C to D: The object moves at a FASTER constant speed from 8 to 12 hours.
D to E: The object stops for 2 hours.
E to F: The object moves at constant speed, for 6 hours, back TOWARDS it's starting point.
- 5Interpret this speed-time graph.0 to 20 s: The object accelerates at a constant rate.
20 to 35 s: The object travels at a CONSTANT SPEED.
35 to 55 s: The object DECELERATES at a constant rate.
55 to 60 s: The object has stopped moving.
- 6A cyclist increases her speed from 2 m/s to 20 m/s in 6 seconds. What is her acceleration?a = (v - u) ÷ t
a = (20 - 2) ÷ 6
3 m/s2
- 7A car speeds up from 12 to 36 m/s with an acceleration of 2 m/s2. How much time did this take?t = (v - u) ÷ a
t = (36 - 12) ÷ 2
t = 12 s - 8Many road accidents happen because drivers reaction times are reduced. Suggest 2 things which can reduce a drivers reaction time.Any 2 from:
- Tiredness / lack of sleep
- Poor visibility.
- Too much alcohol
- Too many drugs. - 9Stopping distance = Thinking distance + Braking distance A car driver sees a red traffic light ahead. She brakes and stops in a total distance of 52 metres. If her thinking distance was 15 metres, what was her braking distance?Stopping distance =
Thinking distance + Braking distance
52 = 15 + Braking distance
Therefore 52 - 15 = Braking distance
Braking distance = 37m - 10Stopping distance = Thinking distance + Braking distance A lorry travels 28 m when stopping from a speed of 4 m/s. If his braking distance was 18 m, what was the driver’s reaction time?Stopping distance =
Thinking distance + Braking distance
28 m = Thinking distance + 18 m
Therefore Thinking distance = 28 - 18
Thinking distance - 10m
Time = distance ÷ speed
Time = 10 ÷ 4
Reaction time = 2.5 s
Informações
- Descrição
- This topic introduces distance, speed, velocity, acceleration and their definitions. as well as the interpretation of distance-time and velocity-time graphs to determine the acceleration of a moving body, and the distance the body travels in a given time. These principles are then applied to the safe stopping distances of vehicles and the factors upon which this depends.
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- Mr S Lee
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