Question

The velocity function (in meters per second) is given for a particle moving along a line.v(t) = 3t − 8, 0 ≤ t ≤ 3(a) Find the displacement.(b) Find the distance traveled by the particle during the given time interval.

Answer 1

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,

$\int\limits^3_0{v(t)} \, dt$ = $\int\limits^3_0 {3t - 8} \, dt$

or

Displacement  =$[\frac{3}{2}t^2-8t]_0^3$

or

Displacement  = $[\frac{3}{2}(3)^2-8(3)]-0$

= - 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 = $\frac{8}{3}$

the velocity changed it direction

thus,

we have the interval for speed as t ∈ $[0, \frac{8}{3}]\ to\ [\frac{8}{3},0]$

therefore,

total distance =  $\int\limits^{\frac{8}{3}}_0 {3t - 8} \, dt + \int\limits^3_{\frac{8}{3}} {3t - 8} \, dt$[/tex]

= $[\frac{3}{2}t^2-8t]_0^{\frac{8}{3}} +[\frac{3}{2}t^2-8t]^3_{\frac{8}{3}}$

=  $[\frac{3}{2}(\frac{8}{3})^2-8(\frac{8}{3}) - (0)] + [\frac{3}{2}(3)^2-8(3)] - [\frac{3}{2}(\frac{8}{3})^2-8(\frac{8}{3})]$

= $\frac{32}{3}-\frac{21}{2}+\frac{32}{3}$

= $\frac{65}{6}$

= 10.83 m