Kinematics: speed of particle

[Laksha]

The speed v m/s of a particle travelling from A to B, at time t s after leaving A, is given by v=10t-t^2. The particle starts from rest at A and comes to rest at B. Show that the particle has a speed of 5m/s or greater for exactly 4√5s.(nov02 no.3)

[Open Intelligence]

v = 10t - t²

v ≥ 5

=> 10t - t² ≥ 5

-t² + 10t - 5 ≥ 0

t² - 10t + 5 ≤ 0

Solving the inequality using [-b±sqrt(b² - 4ac)]/2a

[5 - 4sqrt(5)]/2 ≤ t ≤ [5 + 4sqrt(5)]/2

time interval  = [5 + 4sqrt(5)]/2 - [5 - 4sqrt(5)]/2

                    = 4sqrt(5)