Find the displacement of a particle moving in a straight line with velocity $v(t)=34 t-8 \mathrm{~m} / \mathrm{s}$ over the time interval $[3,5]$.
(Give your answer as a whole or exact number.)
displacement: $\mathrm{m}$

Step 1 ：The displacement of a particle moving in a straight line is given by the integral of the velocity function over the given time interval.

Step 2 ：The velocity function is given as \(v(t) = 34t - 8\) m/s.

Step 3 ：We need to find the displacement over the time interval \([3,5]\).

Step 4 ：This is given by the definite integral of the velocity function from 3 to 5: \(\int_{3}^{5} v(t) dt = \int_{3}^{5} (34t - 8) dt\).

Step 5 ：This integral can be solved by using the power rule for integration, which states that the integral of \(t^n\) is \((1/n+1)t^{n+1}\), and the fact that the integral of a constant is just the constant times the variable of integration.

Step 6 ：So, \(\int_{3}^{5} (34t - 8) dt = [17t^2 - 8t]_{3}^{5}\).

Step 7 ：Now, we substitute the upper and lower limits of the integral: \((17*5^2 - 8*5) - (17*3^2 - 8*3)\).

Step 8 ：This simplifies to \((425 - 40) - (153 - 24)\).

Step 9 ：Further simplification gives \(385 - 129\).

Step 10 ：So, the displacement of the particle over the time interval \([3,5]\) is \(\boxed{256}\) meters.