Question

Cyclist A is moving at 20.0 m/s whereas cyclist B is moving at 12.0 m/s in the same direction and is initially ahead ofA. When they are abreast, they both start to accelerate. Twelve seconds later, B overtakes A when B's speed is 36 m/s. What is A's speed at this point?

Answer 1

Answer:

$v_{fA} = 28 \frac{m}{s}$

Explanation:

The kinematic parameters from the moment the two cyclists begin to accelerate until they meet are:

Initial parameters:

$v_{oA} = 20 m/s$ :  Initial speed of cyclist A

$v_{oB} = 12 m/s$ : Initial speed of cyclist B

Final parameters:

$v_{fB} = 20 m/s$ : Final speed of cyclist B

$d_{A} = d_{B}$

distance of cyclist A = distance of cyclist B

$t_{A} =t_{B} = 12s$

time of cyclist A = time of cyclist B

Cyclist B Kinematics

$v_{fB} = v_{oB} +a_{B} *t$

$36= 12 + a_{B} *12$

$a_{B} = \frac{36-12}{12}$

$a_{B} = 2 \frac{m}{s^{2} }$

$d_{B} =( v_{oB})*( t)+( \frac{1}{2} )*(a_{B})* (t)^{2}$

$d_{B} =( 12*(12)+( \frac{1}{2} )*(2)* (144)}$

$d_{B} = 288m$

Cyclist A Kinematics

$d_{A} =( v_{oA})*( t)+( \frac{1}{2} )*(a_{A})* (t)^{2}$

$d_{A} =(20*( 12)+( \frac{1}{2} )*(a_{A})* 144}$

$d_{A} = 240 + 72*(a_{A})$

$288=240+72*(a_{A} )$

$a_{A} = \frac{288-240}{72}$

$a_{A} = 0.67 \frac{m}{s^{2} }$

$v_{fA} = v_{oA} + a_{A} * t$

$v_{fA} = 20 + 0.67*12$

$v_{fA} = 28 \frac{m}{s}$