Question #276895

A particle moves along x-axis according to expression x=3t²-2t where x is in metre and t in seconds.



a) What is the displacement of the particles in the time interval



t=0 to t=1



t=1 to t=4




b) Calculate the average velocity for the time intervals in (a)



c) Find the instantenous velocity for a particle at 3 seconds

1
Expert's answer
2021-12-07T20:11:37-0500

Given:

x(t)=3t22tx(t)=3t²-2t


a) the displacement of the particles


d01=x(1)x(0)=312210=1md_{01}=x(1)-x(0)=3*1^2-2*1-0=1\:\rm m

d14=x(4)x(1)=(34224)(31221)=39md_{14}=x(4)-x(1)\\=(3*4^2-2*4)-(3*1^2-2*1)=39\:\rm m

b) the average velocity of the particle

v01=d01Δt=1m1s=1m/sv_{01}=\frac{d_{01}}{\Delta t}=\rm\frac{1\: m}{1\: s}=1\: m/s

v14=d14Δt=39m3s=13m/sv_{14}=\frac{d_{14}}{\Delta t}=\rm\frac{39\: m}{3\: s}=13\: m/s

c) the instantaneous velocity


v(t)=x(t)=(3t22t)=6t2v(t)=x'(t)=(3t^2-2t)'=6t-2

at the instant t=3st=3\:\rm s

v(3)=632=16m/sv(3)=6*3-2=16\:\rm m/s


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