COMPOUND MEASUREMENTS - ANSWERS
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1. At the end of a rally the cars which took first and last place had average speeds of 104 km/h and 96 km/h. The first car took 3 hours 52 minutes to complete the course and the last car took 4 hours 5 minutes to complete the course. Approximately how many kilometres was the last car behind the first car, when the first car crossed the finishing line? |
Average Speed = Total Distance Travelled Hence, total distance travelled = Average Speed x Total Time taken. For first place, distance = 104 x
3 52/60. At the time the first car passed the finishing line, Hence the approximate distance between the two cars, |
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2. A car travels along the motorway at an average speed of 112 km/h for 3 hours. The car then travels along minor roads with an average speed of 65 km/h for 2 hours. What is the average speed of the car over the whole journey? |
Total Average Speed = 112 + 65 = 177 = 88.5 km/h |
| 3. The diameter of a car wheel is 56cm. When the car is travelling at 110 km/hour, how many revolutions does the wheel make in one minute? |
The wheel does 1 revolution (full turn) when the circumference of the circle which the wheel makes is turned once. Circumference = 2pr (where r = the radius of the circle and p is approximately 3.14159. The diameter of a circle = 2r, So, Circumference = pd = 3.14159 x 56 = 175.93 (to 2 d.p.) NOTE: This intermediate answer is given to 2 d.p. to make the calculation easier to see. When you do the calculation, make sure that you use all digits until you get to the final answer which is the only number which should be rounded. Now you need to do some comversions: 110 km/hour = 110/60 km/minute (there are 60 minutes in an hour) = 1.83333.... km/minute 1.833333... km/minute = 1,000 x 1.83333..m/minute (as there are 1,000 metres in a kilometre) = 1,833.3333.. (or 1,833 and 1/3) m/minute. If we now convert our circumference to metres by dividing by 100 (100 centimetres in a metre) we get 1.7593 m So, we now have the distance around the wheel as being 1.7593 m and the total distance travelled as being 1,833 and 1/3 m. Divide the total distance travelled by the distance around the wheel and you get the revolutions per minute ie. 1,042.
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4. A long distance lorry driver sets off on a journey with a full tank of diesel (360 litres). When the driver returns to the depot the lorry has travelled 4,638 kilometres. The level of diesel remaining in the tank is checked and the tank is found to have a quarter of a tank of diesel remaining. a) How much diesel was used on the journey? b) What was the average diesel consumption? c) Assuming that the average diesel consumption does not change, how much further can the lorry travel before running out of diesel? |
a) If a 1/4 of a tank of diesel remains then 3/4 has been used i.e. 360 x 3/4 = 270 litres. b) Consumption can be measured in miles per gallon or in this case it is more appropriate to use km per litre. = 4,638 = 17.18 km/litre. c) The lorry has 90 litres of diesel left. If the lorry can do 17.18 km/litre then the lorry can be expected to run for 17.18 x 90 = 1,546.2 km. |
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5. The diagram below shows a prism which has a mass of 3 kg.
a) Calculate the volume of the prism. b) Calculate the density of the prism in g/cm3 of the material from which it is made. |
a) The volume of a prism = Area of Cross-section x length. Calculate the area of the cross-section by splitting it into 2 triangles and a square. Area of a triangle = 1/2 x
base x height.
As the two triangle are the same we can multiply the area of one of
them by 2: (the 1.2 comes from 8.6 - 6.2 divided by 2) Volume = 6.48 x 10 = 64.8 cm3. b) Density = Mass Density = 3 / 64.8 = 0.0463 g/cm3. |
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6. A saleswoman travelled by car to see a customer 200 miles away. She set off from home at 8.00am and arrived at the customer's office at noon. a) What was the car's average speed on the outbound journey? b) The petrol in the car cost 0.82p per litre. The car on average will do 35 miles to the gallon. What is the cost of petrol for the whole journey (i.e. outbound and return)? Assume 1 gallon = 4.5 litres |
a) Total outbound Journey time = 12:00 - 8:00 = 4 hours. Total outbound distance = 200 miles. Average Speed = Total Distance Travelled Average speed = 200 / 4 = 50 mph (miles per hour). b) For both the outbound and return journeys the total distance = 400 miles. No. of gallons used = 400 / 35 = 11.43 (to 2 d.p.) Converting gallons to litres (multiply by 4.5) this is 51.43 litres. Hence the cost = 51.43 x 0.82 = £42.17. |
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7. Greg drives 90 miles to see his grandparents. The distance-time
graph below shows this journey. a) Explain what you think may have happened in the time up to 1 hour and just after 2 hours. b) Calculate Greg's average speed over the whole journey. c) What was Greg's average speed on the return journey?
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a) Only 5 miles is covered in 1 hour, this must be due to a hold
up e.g. an accident. No distance is covered in half an hour, this must mean that Greg has stopped, e.g. for a break at a service station. b) The average speed only really counts when the car is actually moving hence the 3.5 hours he stayed at his grandparents + the half hour at the service station are subtracted from the total time taken = 5 hours. Total distance there and back = 90 + 90 miles Average speed = 180 / 5 = 36 mph. c) On the return journey time = 1.5 hours and the distance = 90 miles. Hence, average speed = 90 / 1.5 = 60 mph |
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8. The diagram below shows the speed-time graph of a non-express
train on its journey from Abbots End station to Newtown station. a) What is the distance between the two stations? b) What was the acceleration of the train before it achieved a constant speed? c) At what rate did the train decelerate before stopping at Newtown station? |
a) The area under the graph is the distance between the two stations = area of triangle + area of square + area of another triangle. = (1.5. x 2) + (5 x 2) + (0.5 x 2) = 14 km b) Acceleration = Speed Acceleration = 2 / 3 km/min2 c) Deceleration = negative acceleration = - 2 / 1 = -2 km/min2.
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