The distance traveled by a particle with velocity v(t) in the interval [3,13] is given by the following integral:
[tex]\int_3^{13}v(t)dt[/tex]However, we can't calculate this integral since we don't have the equation of v(t). What we do have is a table with the value of v for several t's. We can use the values in this table to approximate the integral I wrote above with a trapezoidal sum.
For the function v(t) we have:
[tex]\int_3^{13}v(t)dt=\frac{1}{2}\sum_{n\mathop{=}1}^4{}\lbrack v(t_{n-1})+v(t_n)\rbrack\cdot(\Delta t)_n[/tex]Where the t's with subindices are the t values given by the table:
[tex]\begin{gathered} t_0=3 \\ t_1=7 \\ t_2=10 \\ t_3=12 \\ t_4=13 \end{gathered}[/tex]And the Δt's are the lengths of the intervals defined by two consecutive t's. Let's find each Δt:
[tex]\begin{gathered} (\Delta t)_n=t_n-t_{n-1} \\ (\Delta t)_1=7-3=4 \\ (\Delta t)_2=10-7=3 \\ (\Delta t)_3=12-10=2 \\ (\Delta t)_4=13-12=1 \end{gathered}[/tex]Now I'm going to write the full expression of the trapezoidal sum (expanding the sum):
[tex]\frac{1}{2}\lbrace\lbrack v(t_0)+v(t_1)\rbrack(\Delta t)_1+\lbrack v(t_1)+v(t_2)\rbrack(\Delta t)_2+\lbrack v(t_2)+v(t_3)\rbrack(\Delta t)_3+\lbrack v(t_3)+v(t_4)\rbrack(\Delta t)_4\rbrace[/tex]Now we use the corresponding v(t) values from the table and we obtain:
[tex]\begin{gathered} \frac{1}{2}\lbrack(5+9)\cdot4+(9+14)\cdot3+(14+18)\cdot2+18+24\rbrack=\frac{1}{2}\cdot(14\cdot4+23\cdot3+32\cdot2+18+24) \\ =\frac{1}{2}\cdot231=115.5\approx\int_3^{13}v(t)dt \end{gathered}[/tex]AnswerThen the answer is the third option, 115.50.
