Answer:
[tex]v_{fA} = 28 \frac{m}{s}[/tex]
Explanation:
The kinematic parameters from the moment the two cyclists begin to accelerate until they meet are:
Initial parameters:
[tex]v_{oA} = 20 m/s[/tex] : Initial speed of cyclist A
[tex]v_{oB} = 12 m/s[/tex] : Initial speed of cyclist B
Final parameters:
[tex]v_{fB} = 20 m/s[/tex] : Final speed of cyclist B
[tex]d_{A} = d_{B}[/tex]
distance of cyclist A = distance of cyclist B
[tex]t_{A} =t_{B} = 12s[/tex]
time of cyclist A = time of cyclist B
Cyclist B Kinematics
[tex]v_{fB} = v_{oB} +a_{B} *t\\[/tex]
[tex]36= 12 + a_{B} *12[/tex]
[tex]a_{B} = \frac{36-12}{12}[/tex]
[tex]a_{B} = 2 \frac{m}{s^{2} }[/tex]
[tex]d_{B} =( v_{oB})*( t)+( \frac{1}{2} )*(a_{B})* (t)^{2}[/tex]
[tex]d_{B} =( 12*(12)+( \frac{1}{2} )*(2)* (144)}[/tex]
[tex]d_{B} = 288m[/tex]
Cyclist A Kinematics
[tex]d_{A} =( v_{oA})*( t)+( \frac{1}{2} )*(a_{A})* (t)^{2}[/tex]
[tex]d_{A} =(20*( 12)+( \frac{1}{2} )*(a_{A})* 144}[/tex]
[tex]d_{A} = 240 + 72*(a_{A})[/tex]
[tex]288=240+72*(a_{A} )[/tex]
[tex]a_{A} = \frac{288-240}{72}[/tex]
[tex]a_{A} = 0.67 \frac{m}{s^{2} }[/tex]
[tex]v_{fA} = v_{oA} + a_{A} * t[/tex]
[tex]v_{fA} = 20 + 0.67*12 [/tex]
[tex]v_{fA} = 28 \frac{m}{s}[/tex]
