## Question:

The acceleration of a particle along a straight line is defined by $a = (2t - 9) m/{ s }^{ 2 }$, where t is in seconds. At t = 0, s = 1 m and v = 10 m/s. When t = 9 s, determine (a) the particle’s position, (b) the total distance traveled, and (c) the velocity.

## Step-by-step

$a = 2t – 9 \\ \int _{ 10 }^{ v }{ dv = \int _{ 0 }^{ 1 }{ (2t – 9) dt } } \\ v -10 = { t }^{ 2 } -9t \\ v = { t }^{ 2 } -9t + 10 \\ int_{ 1 }^{ s } ds = \int_{ 0 }^{ t }({ t }^{ 2 } – 9t + 10)dt \\ s – 1 = \frac { 1 } { 3 } { t }^{ 3 } – 4.5{ t }^{ 2 } + 10t \\ s = \frac { 1 } { 3 } { t }^{ 3 } – 4.5 { t }^{ 2 } + 10 t + 1$

Note when $v = { t }^{ 2 } – 9 t + 10 = 0$:

$t = 1.298 s \text{ and } t = 7.701 s$

When $t = 1.298 s,\quad s = 7.13 m$

When $t = 7.701 s,\quad s = -36.63 m$

When $t = 9 s,\quad s = -30.5 m$

\text{ (a) } \quad s = -30.5m \\ \begin{aligned} \text{ (b) } \quad &{ s }_{ To t } = (7.13 – 1) + 7.13 + 36.63 + (36.63 – 30.50) \\ & { s }_{ To t } = 56.0m \end{aligned} \\ \text{ (c) } \quad v = 10\space m/s