Average speed twice the average velocity

A particle leaves a fixed point \(O\) with velocity \(v=10-2t\) m/s, where \(t\) is the time in seconds.

1. Show that, for \(T>5\), the total distance travelled in the first \(T\) seconds is \(T^2-10T+50\) m.[3]
2. Explain why, for \(0<T\le5\), the average speed over the first \(T\) seconds equals the magnitude of the average velocity.[1]
3. Find the exact values of \(T\) for which the average speed over the first \(T\) seconds is twice the magnitude of the average velocity.[4]