The position of a particle moving along the x axis varies in time according to the expression x = 4t 2, where x is in meters and t is in seconds. evaluate its position at the following times.

(a)t=2.10s


(b)t=2.10s +δt
xf=

(c)evaluate the limit of δx/δt as δt approaches zero to find the velocity at t = 2.10 s.
/s

R:  For b) you just need to plug in "3.5 + Δt" as "t" in the formula for x, and simplify: 

x = 2(3.5 + Δt)^2 
x = 2(12.25 + 7Δt + Δt^2) 
x = 2Δt^2 + 14Δt + 24.5 

For c) we need to figure out what Δx is; presumably Δx is the difference in position between t = 3.5 and t = 3.5 + Δt. In other words, Δx = 2Δt^2 + 14Δt. 

So Δx/Δt is therefore 2Δt + 14. The limit of this as Δt approaches zero is 14, so the velocity is 14 at t = 3.5. 

We can check this by taking the derivative of x with respect to t: 

dx/dt = 4t 

So at t = 3.5, dx/dt (or velocity) is equal to 14, which matches what we came up with above using the limit.

It should be d if not then c