Question

From t = 0 to t = 4.70 min, a man stands still, and from t = 4.70 min to t = 9.40 min, he walks briskly in a straight line at a constant speed of 2.95 m/s. What is his vavg in the time interval 2.00 min to 6.70 min?

Answer 1

Task:

From $t = 0$ to $t = 4.70$ min, a man stands still, and from $t = 4.70$ min (282 s) to $t = 9.40$ min, he walks briskly in a straight line at a constant speed of $2.95 \, \text{m/s}$. What is his average speed in the time interval $2.00$ min to $6.70$ min?

Solution:

$$v_{avg} = \frac{s}{t}$$

The average speed in the time interval 2.00 min (120 s) to 6.70 min (402 s):

$$v_{avg} = \frac{s(402 \, s) - s(120 \, s)}{402 \, s - 120 \, s}$$ $$s(402 \, s) = v_{const} \cdot (402 \, s - 282 \, s) = 2.95 \, \frac{m}{s} \cdot 120 \, s = 354 \, m$$ $$s(120 \, s) = 0 \, m$$ $$v_{avg} = \frac{s(402 \, s) - s(120 \, s)}{402 \, s - 120 \, s} = \frac{354 \, m}{282 \, s} = 1.255 \, \frac{m}{s}$$

Answer:

$$v_{avg} = 1.255 \, \frac{m}{s}$$