Answer with Step-by-step explanation:
We are given that a velocity function v(t)=3t-10 for [tex]0\leq t\geq 5[/tex]
a.We have to find the displacement traveled by the particle during the time interval
Displacement =Velocity multiply by time
[tex]ds=v\cdot dt[/tex]
Displacement= S==[tex]\int_{5}^{0}(3t-10)dt[/tex]
S=[tex][3\frac{t^2}{2}-10t]^5_0[/tex]
S=[tex]\frac{3}{2}(5)^2-10(5)[/tex]
[tex]S=\frac{75}{2}-50=\frac{75-100}{2}[/tex]
[tex]S=-\frac{25}{2}=-12.5 meters[/tex]
b.We have to find the distance traveled by the particle during the time interval.
Substitute 3t-10=0
[tex]3t=10[/tex]
[tex]t=\frac{10}{3}[/tex]
Distance [tex]=d_2=\mid \int_{0}^{\frac{10}{3}}(3t-10)dt \mid +\mid \int_{\frac{10}{3}}^{5}(3t-10)dt \mid [/tex]
Distance=[tex]d_2=\mid [\frac{3}{2}t^2-10t}]^{\frac{10}{3}}_0 \mid +\mid [\frac{3}{2}t^2-10t]^5_{\frac{10}{3}}\mid [/tex]
Distance=[tex]d_2=\mid \frac{300}{18}-\frac{100}{3}\mid +\mid \frac{75}{2}-50-\frac{300}{18}+\frac{100}{3} \mid [/tex]
Distance=[tex]d_2=\mid \frac{-100}{6}\mid +\mid \frac{25}{6}\mid [/tex]
Distance=[tex]d_2=\frac{100}{6}+\frac{25}{6}[/tex]
Distance=[tex]d_2=\frac{100+25}{6}=\frac{125}{6}=20.83 meters[/tex]
Hence,the distance traveled by the particle during the time interval =20.83 meters
