Every 2 seconds, Esther takes 5 steps and each of her steps is 0.4 m long.
Therefore, in 2 seconds Esther travels a distance of 5×0.4=2 m.
Solution 1
Every 2 seconds, Paul takes 5 steps and each of his steps is 1.2 m long.
Therefore, every 2 seconds Paul travels a distance of 5×1.2=6 m.
Since Paul travels 6 m every 2 seconds, his speed is 6÷2=3 m/s.
Solution 2
Paul takes 5 steps every 2 seconds and so Paul takes 2.5 steps every second.
The length of each of Paul’s steps is 1.2 m, and so every second Paul travels a distance of 2.5×1.2=3 m.
Thus, Paul travels at a speed of 3 m/s.
Solution 1
Paul travels at a speed of 3 m/s, and so in 120 seconds (2 minutes), Paul travels a distance of 3×120=360 m.
Esther travels 2 m every 2 seconds and so she travels at a speed of 1 m/s.
In 120 seconds, Esther travels a distance of 1×120=120 m.
If Paul and Esther start a race at the same time, then after 2 minutes, Paul will be 360−120=240 m ahead of Esther.
Solution 2
Each of Paul’s steps is 1.2−0.4=0.8 m longer than each of Esther’s steps.
Every 2 seconds, Paul and Esther each take 5 steps, and so in 2×60=120 seconds (2 minutes), Paul and Esther each take 5×60=300 steps.
If Paul and Esther start a race at the same time, then after 2 minutes Paul will be 300×0.8=240 m ahead of Esther.
Solution 3
Paul travels at a speed of 3 m/s, and Esther travels 2 m every 2 seconds, so she travels at a speed of 1 m/s.
This means that in 1 second, Paul travels 3 m and Esther travels 1 m.
Every second, Paul travels 3−1=2 m farther than Esther.
If Paul and Esther start a race at the same time, then after 120 seconds (2 minutes), Paul will be 120×2=240 m ahead of Esther.
Solution 1
Esther travels 2 m every 2 seconds or 1 m/s, and so in 180 seconds (3 minutes), Esther travels 180×1=180 m.
Paul travels at a speed of 3 m/s, which is 2 m/s faster than the speed at which Esther travels.
Therefore, every second, Paul travels 2 m farther than Esther travels.
Since Esther begins the race 180 m ahead of Paul, it will take Paul 180÷2=90 s to catch Esther.
Solution 2
Esther travels 2 m every 2 seconds or 1 m/s, and so in 180 seconds (3 minutes), Esther travels 180×1=180 m.
Each of Paul’s steps is 1.2−0.4=0.8 m longer than each of Esther’s steps.
Since Paul and Esther each step at the same rate (5 steps every 2 seconds), then it will take Paul 180÷0.8=225 steps to catch Esther.
Paul takes 5 steps every 2 seconds, and so it will take Paul 5225×2=45×2=90 s to catch Esther.