Find the displacement of a particle moving in a straight line with velocity $v(t)=34t-8\mathrm{m}/\mathrm{s}$ 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: $(17\ast 5^2-8\ast 5)-(17\ast 3^2-8\ast 3)$.

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 $\boxed{{\displaystyle 256}}$ meters.