# Chapter 2 - Motion Along a Straight Line - Problems - Exercises - Page 62: 2.55

(a) The sprinter has run 10 meters when he reaches his maximum speed. (b) (i) The average velocity is 8.3 m/s (ii) The average velocity is 9.1 m/s (iii) The average velocity is 9.5 m/s

#### Work Step by Step

(a) We can find the acceleration of the sprinter. $a = \frac{v-v_0}{t} = \frac{10~m/s-0}{2.0~s} = 5.0~m/s^2$ We can find the distance covered in the first 2.0 seconds. $x = \frac{1}{2}at^2 = \frac{1}{2}(5.0~m/s^2)(2.0~s)^2$ $x = 10~m$ The sprinter has run 10 meters when he reaches his maximum speed. (b) (i) average velocity = $\frac{\Delta x}{\Delta t}$ We can find the time $t_2$ to run from 10.0 m to 50.0 m. $t_2 = \frac{d}{v} = \frac{40.0~m}{10~m/s} = 4.0~s$ $\frac{\Delta x}{\Delta t} = \frac{50.0~m}{6.0~s} = 8.3~m/s$ (ii) average velocity = $\frac{\Delta x}{\Delta t}$ We can find the time $t_2$ to run from 10.0 m to 100.0 m. $t_2 = \frac{d}{v} = \frac{90.0~m}{10~m/s} = 9.0~s$ $\frac{\Delta x}{\Delta t} = \frac{100.0~m}{11.0~s} = 9.1~m/s$ (iii) average velocity = $\frac{\Delta x}{\Delta t}$ We can find the time $t_2$ to run from 10.0 m to 200.0 m. $t_2 = \frac{d}{v} = \frac{190.0~m}{10~m/s} = 19.0~s$ $\frac{\Delta x}{\Delta t} = \frac{200.0~m}{21.0~s} = 9.5~m/s$

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