Answer
$(a)\space c=0.11\space m/s^{4}$
$(b)\space 3.94\space m/s$
$(c)\space -2.75\space m/s^{2}$
Work Step by Step
Given that, $x=b\space t^{2}-c\space t^{4}-(1)$ ; $Where,\space x= position,\space t= time,\space b=1.82\space m/s^{2}$
(a) In here, they mentioned, x = 0 when t = 2.54 s
So, let's plug these values into equation (1),
$x=b\space t^{2}-c\space t^{4}$
$0=1.82\space m/s^{2}\times(2.54\space s)^{2}- c\times (2.54)^{4}$
$0= 4.62\space m - 41.62\space c\space s^{4}$
$c= \frac{4.62}{41.62}\space m/s^{4}$
(b) To find the velocity, we have to differentiate the equation (1) by t
$x=b\space t^{2}-c\space t^{4}$
$\frac{d}{dt}x=\frac{d}{dt}bt^{2}- \frac{d}{dt}ct^{4}$
$V=b\times2t-c\times3t^{3}-(2)$
Now let's plug known values into equation (2)
$V=1.84\space m/s^{2}\times2\times2.54\space s- 0.11\space m/s^{4}\times3\times(2.54\space s)^{3}$
$V= 9.35\space m/s-5.41\space m/s $
$V=3.94\space m/s-Velocity$
(c) To find the acceleration, we have to differentiate the equation (2) by t
$V=2bt-3ct^{3}$
$\frac{d}{dt}V=\frac{d}{dt}2bt- \frac{d}{dt}3ct^{3}$
$a=2b-9ct^{2}-(3)$
Now let's plug known values into equation (3)
$a=2\times1.82\space m/s^{2}-9\times0.11\space m/s^{4}(2.54\space s)^{2}$
$a=3.64\space m/s^{2}-6.39\space m/s^{2}$
$a=-2.75\space m/s^{2}- Acceleration$
