A jogger runs in a straight line with an average velocity of 4.7 m/s for 3.3 min, and then with an average velocity of 5.8 m/s for 4.4 min. What is her average velocity?

Apr 13, 2017

5.32857142857/sec) = color(red)("average velocity")

Explanation:

Take ${v}_{1}$ as 4.7m/s
${t}_{1}$ = 3.3m

Use this equation to calculate the distance

${v}_{1} \times {t}_{1} = \frac{d}{\cancel{t}} \cdot \cancel{{t}_{1}}$

${v}_{1} \times {t}_{1} = d$

Where dis distance
t is time
v is velocity

Plug in the variables
3.3min must be converted to secs

$3.3 \cancel{\min} \times \frac{60 \sec}{\cancel{\min}} = 198 \sec$

$\frac{4.7 m}{\cancel{\sec}} \times 198 \cancel{\sec} = 930.6 m = {d}_{1}$

Do the same for ${v}_{2}$ that is the another velocity

= $\frac{5.8 m}{\sec} \times \left(4.4 \min \times \frac{60 \sec}{1 \cancel{\min}}\right)$

= $\frac{5.8 m}{\cancel{\sec}} \times 264 \cancel{\sec} = 1531.2 m = {d}_{2}$

$\frac{\Delta d}{\Delta t} = \text{average velocity}$

$\Delta d = {d}_{1} + {d}_{2} = 930.6 m + 1531.2 m = 2461.8$

$\Delta t = {t}_{1} + {t}_{2} = 198 \sec + 264 \sec = 462 \sec$

$\frac{\Delta d}{\Delta t} = \frac{2461.8 m}{462 \sec} = \frac{5.32857142857}{\sec}$

5.32857142857/sec) = color(red)("average velocity")