Answer
$ (a) v(t) = 20m/s $ $a(t) = 16m/s^{2} $ (b) The velocity will be moving to the right in $ t1 \lt \frac{5}{3}$ and to the left in $\frac{5}{3}\lt t2 \lt 3 $.
(c) $- \frac{4}{3} $ m/s
Work Step by Step
(a) First we need to differentiate the original equation of position $ x(t) = (t-1)(t-3)^{2} $, since we know that $ x'(t) = v(t) $
$x'(t) = \frac{d}{dt}(t-1)(t-3)^{2} + (t-1)\frac{d}{dt}(t-3)^{2} $
$ x'(t) = v(t) = 3t^{2} -14t +15 $
We know the interval of time goes from $ 0\leq t \leq 5 $, so we can substitute the $ t =5 $ in our equation of velocity.
$ v(5) = 3(5)^{2} -14(5) +15 $
$ v(5) =20 $
For the acceleration, we know that the derivative of the velocity is the acceleration, thus $v'(t) = a(t) $
$ v'(t) = \frac{d}{dt}(3t^{2} -14t +15) $
$ v'(t) = a(t) = 6x-14 $
In the same interval that goes from $ 0\leq t \leq 5 $, we can replace t for 5 seconds in our equation for acceleration.
$ a(5) = 6(5)-14 $
$ a(5) =16 $
(b) To find the intervals where the particle is moving to the right we must find where the velocity is bigger than 0, $ v(t) \geq 0 $.
$ 0 = 3t^{2} -14t +15 $
$ 0 =(3t-5)(x-3) $
$ t1 = \frac{5}{3}$
$t2 = 3 $
Now we find were the velocity is negative by substituting a time value from each interval on the equation:
for $ t1 \lt \frac{5}{3}$
$ v(\frac{2}{3} =(3(\frac{2}{3})-5)((\frac{2}{3})-3) \gt 0 $
for $\frac{5}{3}\lt t2 \lt 3 $
$ v(2) =(3(2)-5)(2)-3) \lt 0 $
According to this information, the velocity will be moving to the right in $ t1 \lt \frac{5}{3}$ and to the left in $\frac{5}{3}\lt t2 \lt 3 $.
(c) To find the velocity where the acceleration is 0, we must first find the time where acceleration is 0. We replace $a(t)$ with $ 0 $ and look for t.
$ 0 = 6(t)-14 $
$ \frac{14}{6} = t $
Now we substitute this time in $ v(t)$
$ v( \frac{14}{6}) = 3( \frac{14}{6})^{2} -14( \frac{14}{6}) +15 $
$ v(\frac{14}{6}) = - \frac{4}{3} $
