How do I find the velocity, position, speed and acceleration when t=1

s is in feet and t is in seconds

s(t) = t^3 - 5t^2 + 3

Firstly, the position function is always the original function. If s(t) is indeed the original function you were given, then simply plug 1 in s(t).

s(1)=1^3-5^2+3
=-1

Second, the derivative of the position function is velocity. Thus take the derivative of s(t); you can represent this with v(t) if you like.

v(t)=3t^2-10t
v(1)=3-10
=-7

Third, the derivative of the velocity function is acceleration, often expressed as a(t).

a(t)=6t-10
a(1)=-4

Lastly, speed is the absolute value of velocity and you may want to use a graphing calculator to solve this.
Hope this helps!

Edit: @Abdulrahman Gamam, speed and velocity are not the same!

(a) When t = 1, we see that the position is:
s(1) = 1 - 5 + 3 = -1 ft.
(In other words, the particle is 1 ft to the left of the origin.)

(b) By differentiating, we see that the velocity at any time t is:
s'(t) = v(t) = 3t^2 - 10t.

At t = 1:
v(1) = 3 - 10 = -7 ft/s.

(c) By taking the absolute value of the answer in (b), we see that the velocity is 7 ft/s.

(d) Take the second derivative and evaluate it at t = 1.

I hope this helps!

velocity and speed are pretty much the same, all you have to do is find the first derivative (velocity), second derivative (acceleration).

first derive (velocity)
3t^2 - 10t

second derive (acceleration)
6t - 10

plug in (1) for t

6(1) - 10 = -4

answer (-4)