Question

Near the end of a marathon foot race, runner A, running with a velocity of 14 miles/hr, trails runner B by 100 ft. At that instant runner B, running with a velocity of 8 miles/hr, begin their "kick" with an acceleration of 0.25 ft/s².

a) At what distance past that point will the runner A pass runner B?

b) At what distance past that same point will runner B pass runner A?

Answer 1

vA = 14 mile/hr = 20.53 ft/s

vB = 11.7 ft/s

aB = 0.25 ft/s^2

after t time A past B

xA = vA*t = 20.53*t

XB -100 = vB*t + (0.5*aB*t^2)

XB = 100 + (11.7t) + (0.5*0.25*t^2)

XA = XB

20.53*t = 100 + (11.7t) + (0.5*0.25*t^2)

t = 14.16 s

at point XA = 20.53*14.16 = 290.7 ft   <<<<=====ANSWER

part(b)

speed of B after t time

vB1 = vB + aB*t = 11.7 + (0.25*14.16) = 15.24 ft/s

let B pass A after t1 time

XA1 = VA*t1 = 20.53*t1

XB1 = (VB1*t1) + (0.5*aB*t1^2)

XB1 = (15.24*t1) + (0.5*0.25*t1^2)

XA1 = XB2

20.53*t1 = (15.24*t1) + (0.5*0.25*t1^2)

t1 = 42.32 s

distance at which B pasts A = 20.53*42.32 = 868.8 ft