Question

a student goes for a walk each day for four days. each day, he records the distance he walks and the time it takes in a table as shown. on which day did the student have the greatest average speed? day 1 day 2 day 3 day 4

Explanation:

Step1: Recall speed formula

The formula for average speed is $v=\frac{d}{t}$, where $v$ is average speed, $d$ is distance, and $t$ is time.

Step2: Calculate speed for Day 1

For Day 1, $d = 100$ m and $t=90$ s. So, $v_1=\frac{100}{90}=\frac{10}{9}\approx1.11$ m/s.

Step3: Calculate speed for Day 2

For Day 2, $d = 150$ m and $t = 140$ s. So, $v_2=\frac{150}{140}=\frac{15}{14}\approx1.07$ m/s.

Step4: Calculate speed for Day 3

For Day 3, $d = 200$ m and $t = 200$ s. So, $v_3=\frac{200}{200}=1$ m/s.

Step5: Calculate speed for Day 4

For Day 4, $d = 250$ m and $t = 235$ s. So, $v_4=\frac{250}{235}=\frac{50}{47}\approx1.06$ m/s.

Step6: Compare speeds

Comparing $v_1\approx1.11$ m/s, $v_2\approx1.07$ m/s, $v_3 = 1$ m/s, and $v_4\approx1.06$ m/s, we can see that $v_1$ is the greatest.

Answer:

Day 1