Answer:
Acceleration of one cyclist=

Yes, the acceleration is higher of another cyclist who accelerates from 0 to 30 m/s in 8 seconds.
Step-by-step explanation:
We are given that
Initial velocity of one cyclist, u=0 m/s
Final velocity of one cyclist, v=8m/s
Time, t=3 s
Initial velocity of another cyclist, u'=0
Final velocity of another cyclist, v'=30m/s
Time, t'=8 s
We know that
Acceleration,

Using the formula

Acceleration of one cyclist=

Acceleration of another cyclist, a'=

Acceleration of another cyclist, a'=

Yes, the acceleration of another cyclist is higher than the cyclist which accelerates from 0m/s to 8m/s.