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

Answer:

a) X = 17.64 m

b) X = 17.64 + 4∆t^2 + 16.8∆t

c) Velocity = lim(∆t→0)⁡〖∆X/∆t〗 = 16.8 m/s

Explanation:

a) The position at t = 2.10s is:  

                 X = 4t^2

                 X = 4(2.10)^2

                 X = 17.64 m

b) The position at t = 2.10 + ∆t s  will be:

                 X = 4(2.10 + ∆t)^2

                 X = 17.64 + 4∆t^2 + 16.8∆t  m

c) ∆X is the difference between position at t = 2.10s and t = 2.10 + ∆t so,

                 ∆X= 4∆t^2 + 16.8∆t

Divide by ∆t on both sides:  

                ∆X/∆t =  4∆t + 16.8

 Taking the limit as ∆t approaches to zero we get:  

              Velocity =lim(∆t→0)⁡〖∆X/∆t〗 = 4(0) + 16.8

              Velocity = 16.8 m/s

 

Answer 2

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.