Please solve 11A and 11B 11. A man with a stopwatch stands perfectly still from t = 0 to t = 5.00 min. From t= 5.00 min to t = 10.00 min, hewalks straight ahead at a speed of 2.20 m/s. What is his average velocity and average acceleration during the follow-ing time intervals:a) 2.00 min to 8.00 minb) 3.00 min to 9.00 min

a) Average velocity is, vavg=∆x∆t=vt∆t=2.20 m/s8 min-5 min8 min-2 min=1.1 m/s Average acceleration…

Answered: 11. A man with a stopwatch stands…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Please solve 11A and 11B

11. A man with a stopwatch stands perfectly still from t = 0 to t = 5.00 min. From t= 5.00 min to t = 10.00 min, he
walks straight ahead at a speed of 2.20 m/s. What is his average velocity and average acceleration during the follow-
ing time intervals:
a) 2.00 min to 8.00 min
b) 3.00 min to 9.00 min

Expert Solution

Step 1

a) Average velocity is,

$\begin{array}{rcl} v_{avg} & = & \frac{∆ x}{∆ t} \\ = & \frac{v t}{∆ t} \\ = & \frac{(2.20 m / s) (8 \min - 5 \min)}{8 \min - 2 \min} \\ = & 1.1 m / s \end{array}$

Average acceleration is,

$\begin{array}{rcl} a_{avg} & = & \frac{∆ v}{∆ t} \\ = & \frac{(2.20 m / s - 0 m / s)}{(8 \min - 2 \min) (\frac{60 s}{1 \min})} \\ = & 6.11 \times 10^{- 3} m / s^{2} \end{array}$

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