General motion in a straight line

Could anyone give me a bit of direction for the following question, please?

In the first part of motion, when t<10, what is the velocity in terms of a at t=10 ?

In the second part of motion, what is the velocity when t= 10 ?

Equate them, they must be equal

I already had an idea for that second part, it is the first part that bothers me. I was going to use the equation but isn't it only for 10 < t < 20

When I say 'part of motion' I am referring to the fact that the particle has two different parts of motion; in the first part, it has constant aceeleration, and in the second one it no longer has constant acceleration.

I was not referring to parts (a) and (b) of the question it self -- because from what you've just said it seems to me this is what you inferred.

Anyway, $v = \dfrac{800}{t^2} - 2$ is for $10 \leq t \leq 20$, so of course you can use $t=10$ here.

I was using your logic actually but when the particle is in constant acceleration, so when t < 10, can I still use $v = \dfrac{800}{t^2} - 2$ is for $10 \leq t \leq 20$?

No you cant use that, and it should be pretty obvious that if you differentiate this you wont get a constant acceleration so using it for motion when 0<t<10 would be in contradiction to what the question tells you.

You need to construct your own velocity equation based on the information and use that for when 0<t<10

Thanks to you, I managed to complete the first two part of the questions. However, I'm stuck for part c. I integrated $v = \dfrac{800}{t^2} - 2$ to become $v = \dfrac{800}{t} - 2t + c$. But I am not sure how I have to proceed from now.