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Ok so i understand the first part of this question:: part a an object is acceler

ID: 2127635 • Letter: O

Question

Ok so  i understand the first part of this question::


part a

an object is accelerating uniformly at 1.00 m/s^2. It starts from rest at time t=0.

that is its velocity at time t=2.00.

how far does it travel from time t=1.90s to t=2.10s?

t = 2.1 sec

s = u*t+ 0.5*a*t^2

u = 0

a = 1 m/s^2

s(2.1) = 2.205 m

s(1.9)= 0.5*1*1.9^2 = 1.805 m

s(2.1) - s(1.9) = 0.4 m is ans


Here is the next part i need help with please:

What is its average velocity during the 0.20s interval? note that this 0.2s interveral surrounds  the 2.00s point of part a.  0.10s on each side.

Explanation / Answer

s = u*t+ 0.5*a*t^2

u = 0

a = 1 m/s^2

s(2.1) = 2.205 m

s(1.9)= 0.5*1*1.9^2 = 1.805 m

s(2.1) - s(1.9) = 0.4 m is ans


Here is the next part i need help with please:

What is its average velocity during the 0.20s interval? note that this 0.2s interveral surrounds  the 2.00s point of part a.  0.10s on each side.