Real-Life and Distance-Time Graphs - Higher

Read up from the time axis and then across to the distance axis.

1. At 10 minutes the graph is at 2 km, so the cyclist is 2 km from home.

Start at the required temperature and read across to the line.

1. The horizontal line through 35\degree\mathrm C meets the graph above 5 minutes.

A horizontal section represents no change in position.

1. The horizontal section lasts from 10 to 20 minutes.
2. The cyclist is stationary for 20-10=10 minutes.

Use distance travelled divided by time taken.

1. The trolley travels 50 m in 20 seconds.
2. Its speed is 50\div20=2.5 m/s.

Consider both the direction of the line and whether its gradient changes.

1. The distance from base decreases at a constant rate.
2. The vehicle therefore travels towards base at constant speed.

Add the distance of each stage, including any return towards home.

1. The three stages cover 5 km, 5-2=3 km and 4-2=2 km.
2. The total distance is 5+3+2=10 km.

The total time includes the stop, and it must be expressed in hours.

1. Average speed = total distance \div total time, and the total time includes the stop.
2. The graph ends at 8 km after 40 minutes, so the total distance is 8 km and the total time is 40 minutes.
3. The speed is wanted in km/h, so change the time to hours: 40\div60=\frac23 of an hour.
4. Average speed =8\div\frac23=8\times\frac32=12 km/h.

Subtract the stationary time before calculating the average speed.

1. The stop lasts 25-15=10 minutes.
2. The moving time is 45-10=35 minutes and the distance is 3.5 km.
3. The average moving speed is 3.5\div(35/60)=6 km/h.

Read the two times and find their difference.

1. Read across from 60\degree\mathrm C to the graph and down: the temperature is 60\degree\mathrm C at 10 minutes.
2. Read across from 30\degree\mathrm C to the graph and down: the temperature is 30\degree\mathrm C at 30 minutes.
3. The time taken is 30-10=20 minutes.

Find the hourly charge from the increase in cost, then include the initial charge.

1. The initial charge is £6.
2. The hourly charge is (22-6)\div4=4 pounds.
3. For 7 hours the cost is 6+7\times4=34 pounds.

Read both temperatures; the sensor reading does not start at zero.

1. Find each reading on the sensor reading axis, read across to the line, then read down to the temperature.
2. A reading of 50 units corresponds to 20\degree\mathrm C.
3. A reading of 90 units corresponds to 40\degree\mathrm C.
4. The temperature increase is 40-20=20\degree\mathrm C.

Find the two times when the vertical gap between the lines is exactly 2 km.

1. At any time, the distance between the walkers is the vertical gap between the two lines.
2. At 10 minutes, A is 1 km and B is 3 km from the gate, a gap of 3-1=2 km. Before this the gap is bigger.
3. At 30 minutes, A is 3 km and B is 1 km from the gate, again a gap of 2 km. After this the gap gets bigger.
4. Between 10 and 30 minutes the gap is less than 2 km (they pass each other at 20 minutes).
5. So they are less than 2 km apart for 30-10=20 minutes.

The crossing gives time since A started, not time since B started.

1. The lines cross at 45 minutes after A starts.
2. B started at 15 minutes, so B has cycled for 45-15=30 minutes.

Find the rate at which the volume falls while the tank drains.

1. The tank drains from 20 minutes, when the line starts to slope down.
2. Between 20 and 30 minutes the volume falls from 60 to 30 litres, so it drains at (60-30)\div10=3 litres per minute.
3. At 30 minutes there are 30 litres left. At 3 litres per minute this takes 30\div3=10 more minutes.
4. The tank is empty 30+10=40 minutes after the start.

Find the tangent’s gradient and compare it with total distance divided by total time.

1. The tangent goes through (2.5,0) and (7.5,100). Its gradient estimates the speed at 5 seconds: 100\div(7.5-2.5)=20 m/s.

2. In the first 5 seconds, the robot travels 50 m, so its average speed is 50\div5=10 m/s.
3. The speed at 5 seconds is about 20-10=10 m/s greater.

Use the total distance there and back and the total time, including the stop.

1. From the graph, the library is 6 km from home and the cyclist leaves it at 45 minutes.
2. The ride home takes time = distance \div speed =6\div12=0.5 hours, which is 30 minutes. So the cyclist is home at 45+30=75 minutes.
3. Total distance =6+6=12 km. Total time =75 minutes =75\div60=1.25 hours.
4. Average speed = total distance \div total time =12\div1.25=9.6 km/h.

Find the distance and the time for that stage, then change kilometres to metres and minutes to seconds.

1. Between 20 and 40 minutes the distance travelled goes from 5 km to 20 km, so the train travels 20-5=15 km in 20 minutes.
2. The speed is wanted in metres per second. 15 km =15\,000 m and 20 minutes =20\times60=1200 seconds.
3. Speed = distance \div time =15\,000\div1200=12.5 m/s.

The downward gradient is the net rate of change, after allowing for the inflow.

1. The volume decreases by 100-40=60 litres in 20 minutes.
2. The net loss is 60\div20=3 litres per minute.
3. The drain must remove the 7 litres entering each minute as well as the net loss: 7+3=10 litres per minute.

Use the magnitude of the tangent’s gradient as the speed, then divide the remaining distance by it.

1. The tangent falls 18 m in 4.5-1.5=3 seconds, so the estimated speed is 6 m/s.
2. At t=3 seconds the robot is 9 m from the dock.
3. At a constant 6 m/s it would take another 9\div6=1.5 seconds.

Find B’s speed before the meeting, then allow for the break and remaining distance.

1. B covers 6 km from t=15 to t=30, so B’s speed is 6\div15=0.4 km per minute.
2. After the meeting, B has 6 km left, taking 6\div0.4=15 minutes of cycling.
3. Including the break, B finishes at 30+5+15=50 minutes. A finishes at 60 minutes.
4. B finishes 60-50=10 minutes before A.