Question

The leader of a bicycle race is traveling with a constant velocity of +11.10 m/s and is 10.0 m ahead of the second-place cyclist. The second-place cyclist has a velocity of +9.50 m/s and an acceleration of +1.20 m/s2. How much time elapses before he catches the leader?

Answer 1

D1 = distance 1 = 11.10 m/s t

D2 = 10m + 11.10 m/s t

D2 = u2 t + 1/2 a2 t^2

u2= speed 2 = 9.50m/s

10m + 11.10 m/s t = 9.50 m/s t + 1/2 (1.20 m/s^2) t^2

Solve for t

10m + 11.10 m/s t = 9.50 m/s t + 0.6 m/s^2 t^2

10 + 11.10 t = 9.5 t + 0.6 t^2

0 = 0.6t^2 - 1.6t - 10

Apply quadratic equation:

`t=\frac{-b\pm\sqrt[\placeholder{⬚}]{b^2-4* a* c}}{2a}`
`t=\frac{1.6\pm\sqrt[\placeholder{⬚}]{(-1.6)^2-4(0.6)(-10)}}{2(0.6)}`
`t=\frac{1.6\pm\sqrt[\placeholder{⬚}]{26.56}}{1.2}`
`t=(1.6+5.15)/(1.2)=5.63\text{ s}`

Answers: 5.63 sec