Answer
(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.
$t_2 = \frac{d}{v} = \frac{190.0~m}{10~m/s} = 19.0~s$
$\frac{\Delta x}{\Delta t} = \frac{200.0~m}{21.0~s} = 9.5~m/s$
