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

`v_(fA) = 28 (m)/(s)`

Step-by-step 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) = (36-12)/(12)`

`a_(B) = 2 (m)/(s^(2) )`

`d_(B) =( v_(oB))*( t)+( (1)/(2) )*(a_(B))* (t)^(2)`

`d_(B) =( 12*(12)+( (1)/(2) )*(2)* (144)}`

`d_(B) = 288m`

Cyclist A Kinematics

`d_(A) =( v_(oA))*( t)+( (1)/(2) )*(a_(A))* (t)^(2)`

`d_(A) =(20*( 12)+( (1)/(2) )*(a_(A))* 144}`

`d_(A) = 240 + 72*(a_(A))`

`288=240+72*(a_(A) )`

`a_(A) = (288-240)/(72)`

`a_(A) = 0.67 (m)/(s^(2) )`

`v_(fA) = v_(oA) + a_(A) * t`

`v_(fA) = 20 + 0.67*12`

`v_(fA) = 28 (m)/(s)`