Question

at a track meet, two runners finish their races at almost the same time. runner a completes the 100 - meter race in 11.2 seconds, while runner b finishes the 110 - meter race in 12.1 seconds. based on the given times and distances, which runner had the higher average speed? a. runner a, with an average speed of 8.93 m/s b. runner b, with an average speed of 8.92 m/s c. runner a, with an average speed of 9.09 m/s d. runner b, with an average speed of 9.09 m/s

Explanation:

Step1: Recall speed formula

Speed $v=\frac{d}{t}$, where $d$ is distance and $t$ is time.

Step2: Calculate Runner A's speed

Runner A runs $d_A = 100$ meters in $t_A=11.2$ seconds. So $v_A=\frac{d_A}{t_A}=\frac{100}{11.2}\approx8.93$ m/s.

Step3: Calculate Runner B's speed

Runner B runs $d_B = 110$ meters in $t_B = 12.1$ seconds. So $v_B=\frac{d_B}{t_B}=\frac{110}{12.1}\approx9.09$ m/s.

Step4: Compare speeds

Since $9.09>8.93$, Runner B has a higher - average speed.

Answer:

D. Runner B, with an average speed of 9.09 m/s