How do you find the total displacement for the particle whose position at time $t$ is given by $s(t)=3t^2-5t$ on the interval $0<=t<=2$ ?

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How do you find the total displacement for the particle whose position at time $t$ is given by $s(t)=3t^2-5t$ on the interval $0<=t<=2$ ?

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Answer 1 · 2015-01-12T21:07:45

The total displacement will be the difference between the final and initial positions.
You have:
Initial position; for $t=0$
$s_i=s(0)=3×0^2-5×0=0$
Final position; for $t=2$
$s_f)=s(2)=3×(2)^2-5×2=12-10=2$

Total displacement $=s_f)-s_i=2-0=2$
(Km, m, cm, mm....)

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How do you find the total displacement for the particle whose position at time $t$ is given by $s(t)=3t^2-5t$ on the interval $0<=t<=2$ ?

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How do you find the total displacement for the particle whose position at time tt is given by s(t)=3t^2-5ts(t)=3t^2-5t on the interval 0<=t<=20<=t<=2?

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How do you find the total displacement for the particle whose position at time $t$ is given by $s(t)=3t^2-5t$ on the interval $0<=t<=2$ ?