Answer:
a. [tex]t=3.44s[/tex]
b. [tex]x=12.45m[/tex]
c. [tex]v_f=7.26\frac{m}{s}[/tex]
Explanation:
The bicyclist's friend moves with constant speed. So, we have:
[tex]x=vt[/tex]
Th bicyclist moves with constant acceleration and starts at rest ([tex]v_0=0[/tex]). So, we have:
[tex]x=v_0t+\frac{at^2}{2}\\x=\frac{at^2}{2}[/tex]
a. When he catches his friend, both travels the same distance, thus:
[tex]vt=\frac{at^2}{2}\\t=\frac{2v}{a}\\t=\frac{2(3.63\frac{m}{s})}{2.11\frac{m}{s^2}}\\t=3.44s[/tex]
b. We can use any of the distance equations, since both travels the same distance:
[tex]x=vt\\x=3.63\frac{m}{s}(3.44s)\\x=12.45m[/tex]
c. The bicyclist final speed is:
[tex]v_f=v_0+at\\v_f=at\\v_f=2.11\frac{m}{s^2}(3.44s)\\v_f=7.26\frac{m}{s}[/tex]
