AS maths question help

Hi, could someone please help with this:

A bird leaves its nest at time t = 0 for a short flight along a straight line. The bird then returns to its nest. The bird is modelled as a particle moving in a straight horizontal line. The distance, s metres, of the bird from its nest at time t seconds is given by s = 1/10 (t^4 – 20t^3 + 100t^2), where 0 ≤ t ≤10 (a) Explain the restriction, 0 ≤ t≤ 10.

I really dont understand this question. could someone help please?

Well, the bird flies away from its nest and comes back. How far away is the bird when $t=0$? So the restriction must mean that...?

well it would be zero metres away? but i dont get what im meant to write for 3 marks.

Hi, could someone please help with this:

A bird leaves its nest at time t = 0 for a short flight along a straight line. The bird then returns to its nest. The bird is modelled as a particle moving in a straight horizontal line. The distance, s metres, of the bird from its nest at time t seconds is given by s = 1/10 (t^4 – 20t^3 + 100t^2), where 0 ≤ t ≤10 (a) Explain the restriction, 0 ≤ t≤ 10.

I really dont understand this question. could someone help please?

Part a:
When t = 0, s = 0 too. That means Displacement is 0 when the bird is at rest.
When t = 10, s = 0 again. That means the bird flew for 10 seconds and came back to the nest in 10 seconds, therefore the displacement is 0 as it returned.
Part b:
S = 1/10(t⁴-20t³+100t²)
Expand the brackets and get s = 1/10t⁴-2t³+10t²
Differentiate S to get v, because the velocity is 0 when the bird first comes to rest.
ds/dt = 2/5t³-6t²+20t
So v = 2/5t³-6t²+20t
Put v = 0:
2/5t³-6t²+20t = 0
t = 0, t = 5 and t = 10
T must be 5 because the other two values of t are when the bird is at rest and when the bird comes back.
Substitute in the value of t = 5 into the S equation:
s = 1/10((5)⁴-20(5)³+100(5)²) = 62.5m
You're welcome 😂❤️