Answer:
a) - 10.5 m
b) 10.83 m
Step-by-step explanation:
Given:
Velocity function , v(t) = 3t - 8, 0 ≤ t ≤ 3
a)
For displacement, integerating the function form time 0 to 3
thus,
[tex]\int\limits^3_0{v(t)} \, dt[/tex] = [tex]\int\limits^3_0 {3t - 8} \, dt[/tex]
or
Displacement =[tex][\frac{3}{2}t^2-8t]_0^3[/tex]
or
Displacement = [tex][\frac{3}{2}(3)^2-8(3)]-0[/tex]
= - 10.5 m
b) For total distance, let us first find the intervals where the velocity went from positive to negative
thus,
3t - 8 = 0
t = [tex]\frac{8}{3}[/tex]
the velocity changed it direction
thus,
we have the interval for speed as t ∈ [tex][0, \frac{8}{3}]\ to\ [\frac{8}{3},0][/tex]
therefore,
total distance = [tex]\int\limits^{\frac{8}{3}}_0 {3t - 8} \, dt + \int\limits^3_{\frac{8}{3}} {3t - 8} \, dt[/tex][/tex]
= [tex][\frac{3}{2}t^2-8t]_0^{\frac{8}{3}} +[\frac{3}{2}t^2-8t]^3_{\frac{8}{3}}[/tex]
= [tex][\frac{3}{2}(\frac{8}{3})^2-8(\frac{8}{3}) - (0)] + [\frac{3}{2}(3)^2-8(3)] - [\frac{3}{2}(\frac{8}{3})^2-8(\frac{8}{3})][/tex]
= [tex]\frac{32}{3}-\frac{21}{2}+\frac{32}{3}[/tex]
= [tex]\frac{65}{6}[/tex]
= 10.83 m