# What is the average velocity of the particle between t=1.00 s and t=3.00 s?

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The position of a particular particle as a function of time is given by r⃗ = ( 9.60t⋅i^ + 8.85j^ – 1.00t2⋅k^)m, where t is in seconds.

Express your answer in terms of the unit vectors i^,j^, and k^.

• When t=1.00s, the position is

r⃗ ₁ = ( 9.60(1.00)î + 8.85ĵ – 1.00(1.00)²k̂ ) m

. . = ( 9.60î + 8.85ĵ – 1.00k̂ ) m

When t=3.00s, the position is

r⃗ ₂ = ( 9.60(3.00)î + 8.85ĵ – 1.00(3.00)²k̂ ) m

. . = ( 28.8î + 8.85ĵ – 9.00k̂ ) m

Displacement Δr⃗ is r⃗ ₂ – r⃗ ₁ and time interval is Δt = (3.00 – 1.00)s = 2.00s

Average veocity v⃗ = Δr⃗ /Δt

= [ ( (28.8-9.60)î + (8.85-8.85)ĵ – (9.00-(-1.00)k̂ ) ] / 2.00 m/s

= [ ( 19.2î + 0ĵ – 8.00k̂ ) ]/2.00 m/s

= ( 9.6î + 0ĵ – 4.00k̂ ) m/s

• r⃗ = ( 9.60t⋅i^ + 8.85j^ – 1.00t2⋅k^)m,

So,

For, t= 1sec, r(1) = ( 9.60⋅i^ + 8.85j^ – 1.00t2⋅k^)m, ………………………… 

and

For, t = 3 sec, r(3) = ( 9.60*3⋅i^ + 8.85j^ – 1.00t2⋅k^)m = ( 28.80⋅i^ + 8.85j^ – 1.00t2⋅k^)m, ……. 

So,

Average velocity = [r(3) – r(10] / (3 – 1) =( 14.40⋅i^ + 0.0j^ – 0.02⋅k^)m/sec. <——————-/

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