Question

A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.1m/s . Two seconds later the bicyclist hops on his bike and accelerates at 2.2m/s2 until he catches his friend.

How much time does it take until he catches his friend (after his friend passes him)?

How far has he traveled in this time?

What is his speed when he catches up?

Answer 1

1 - bicyclist, 2 - friend, v - speed, a - acceleration, x - displacement, t - time

v01 = 0 m/s, a01 = 2.2 m/s^2,
v02 = 3.1 m/s

formula for distance traveled at constant acceleration: a1 = dv1/dt = a01
v1 = dx1/dt = a01*t + v01
x1 = 0.5*a01*t^2 + v01*t+ x01

formula for distance traveled at constant speed: a2 = dv2/dt = 0
v2 = dx2/dt = v02
x2 = v02*t + x02

for time t <= 2, x02 = 0 m,
distance traveled by friend in 2 seconds: x2 = v02*t + x02 = 3.1*2 + 0 = 6.2 m

for time t > 2, x01 = 0 m, x02 = 6.2 m, v01 = 0 m/s,
when caught up x1 = x2: 0.5*a01*t^2 + v01*t+ x01 = v02*t + x02
rearranging: 0.5*a01*t^2 + (v01-v02)*t+ (x01-x02) = 0
solve using quadratic equation: t = { -(v01-v02) +/- sqrt[ (v01-v02)^2 - 2*a01*(x01-x02) ] } / a1
substituting values: t = { 3.1 +/- 6.4 ] } / 2.5 = 3.8 s (only possible answer because alternative gives negative times which is obviously impossible)

it takes approximately an additional 4 seconds on top of the 2 seconds (total: 5.8 s) to catch up with his friend.

distance traveled: x1 = 0.5*a01*t^2 + v01*t + x01 = 0.5*3.1*3.8^2 + 0*3.8 + 0 = 22.4 m

speed at meeting: v1 = a01*t + v01 = 2.5*3.8 + 0 = 9.5 m/s