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

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

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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 $t = 0$ $t=0$
$s_{i} = s (0) = 3 \times 0^{2} - 5 \times 0 = 0$ $s_i=s(0)=3×0^2-5×0=0$
Final position; for $t = 2$ $t=2$
$s_{f}) = s (2) = 3 \times {(2)}^{2} - 5 \times 2 = 12 - 10 = 2$ $s_f)=s(2)=3×(2)^2-5×2=12-10=2$

Total displacement $= s_{f}) - s_{i} = 2 - 0 = 2$ $=s_f)-s_i=2-0=2$
(Km, m, cm, mm....)

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

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