Answer:
c. -t² + 5t < 5
Explanation:
The cyclist starts at an initial elevation of 0 at time t = 0
The cyclist's elevation again becomes 0 when the graph hits the t axis
To solve for when the cyclist reaches elevation of 0 after setting off we can set -t² + 5t = 0 and solve for t
-t² + 5t = 0
=> -t² = - 5t (move 5t to the right)
=> -t² = - 5 (divide both sides by -t)
=> t = 5 (multiply both sides by -1)
So the cyclist reaches an elevation of 0 again at t= 5
So the cyclist is above elevation 0 when
-t² + 5t < 5, t ≥ 0 (since we cannot have negative time)