Answer:
a. [tex]x=540.8m[/tex]
b. [tex]v_f=52\frac{m}{s}[/tex]
Explanation:
a. David is moving with constant speed. While Tina is accelerating, we use the equations of uniformly accelerated motion.
For David we have:
[tex]x=vt[/tex]
For Tina:
[tex]x=v_0t+\frac{at^2}{2}\\v_0=0\\x=\frac{at^2}{2}[/tex]
They both travel the same distance from the moment David passes her until she passes David:
[tex]vt=\frac{at^2}{2}\\t=\frac{2v}{a}\\t=\frac{2(26\frac{m}{s})}{2.5\frac{m}{s^2}}\\t=20.8s[/tex]
Knowing this, we can calculate the distance:
[tex]x=vt\\x=26\frac{m}{s}(20.8s)\\x=540.8m[/tex]
b. To calculate her speed when she passes him, we use:
[tex]v_f=v_0+at\\v_f=at\\v_f=(2.5\frac{m}{s^2})20.8s\\v_f=52\frac{m}{s}[/tex]
