Question

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
______m

(b)t=2.10s +Δt
xf=________m

(c)Evaluate the limit of Δx/Δt as Δt approaches zero to find the velocity at t = 2.10 s.
_______m/s ...?

Answer 1

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.