The position of a particle moving alongxaxis depends on the time according to the equation $\mathit{x}\mathbf{=}\mathit{c}{\mathit{t}}^{\mathbf{2}}\mathbf{‐}\mathit{b}{\mathit{t}}^{\mathbf{3}}$, wherexis in meters and tin seconds. What are the units of –(a)Constant c and (b)Constant b? Let their numerical values be 2.0and 3.0respectively. (c) At what time does the particle reach its maximum positive x position? From$\mathit{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$to $\mathit{t}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$. (d) what distance does the particle move? (e) what is its displacement? Find its velocity at times- (f) $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$(g) $\mathit{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$(h)$\mathit{t}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$(i)$\mathit{t}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$Find its acceleration at times- (j)$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$(k)$\mathit{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$(l)$\mathit{t}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$(m)$\mathit{t}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$.

Units of the unknown constant can be found by comparing the left and right side of the equations. Position, velocity and acceleration values can be found using the relationship between displacement, time, velocity and acceleration.

Find unit of c as left side of the equation represents the length so right side should be the length, so units of c and b are taken such that time unit will be canceled and remaining unit will be only of length.

For maximum coordinate v=0 . Using the givencondition, find time for maximum coordinate.

For given values of time find total distance travelled by particle by putting the values in the given equation.

Find velocity equation by taking derivative of given equation with respect to t and then putting the given times find velocities.

Similarly taking derivative of velocity equation find equation of acceleration and then putting the values of times find value of acceleration.

The formula for velocity at given time is as follow,

$v\left(t\right)=\frac{dx}{dt}$(i)

The equation given in the question is,

$x=c{t}^2-b{t}^3$(ii)

Unit of $c{t}^2$is length then unit of $c$must be $length/t^2$or $m/s^2$.

Since $bt3$has unit of length so b must have unit$lenght/t^3$ or$m/s^3$.

At maximum coordinate velocity of particle is zero. Velocity equation can be found by taking derivative of the given equation.

$$\begin{gathered} v\left(t\right)=\frac{dx}{dt} \\ =\frac{d\left(c{t}^2-b{t}^3\right)}{dt} \\ =2ct-3b{t}^2 \end{gathered}$$

$v=0$at $t=0$. Also $v=0$for $t$which can be found by

role="math" localid="1656407021590" $$\begin{gathered} 0=2ct-3b{t}^2 \\ t=\frac{2c}{3b} \\ =\frac{2\times 3.0}{3\times 2.0} \\ =1.0s \end{gathered}$$

Thus, it takes $\mathrm{t}=1.0\mathrm{s}$to reach maximumx coordinate.

For $t=4s$particle moves from $x=0$to $x=1.0\mathrm{m}$, then turns around and goes back to a distance $x\left(4\mathrm{s}\right)$,

Hence,

$$\begin{gathered} x\left(4\mathrm{s}\right)=\left(\frac{3.0\mathrm{m}}{{\mathrm{s}}^2}\right){\left(4.0\mathrm{s}\right)}^2-\left(\frac{2.0\mathrm{m}}{{\mathrm{s}}^3}\right){\left(4.0\mathrm{s}\right)}^3 \\ =-80\mathrm{m} \end{gathered}$$

Thus, the total path length it travels is$1.0\mathrm{m}+1.0\mathrm{m}+80\mathrm{m}=82\mathrm{m}$.

Displacement is shortest distance between initial and final positions that is

$$\begin{gathered} ∆x=x_2-x_1 \\ {x}_1=0\mathrm{and}{x}_2=-80\mathrm{m}. \end{gathered}$$

Thus,

$∆x=-80\mathrm{m}$

Thus, the displacement of the particle is $-80\mathrm{m}$.

The velocity equation is,

$$\begin{gathered} v=2ct-3b{t}^2 \\ =\left(2\times \frac{3\mathrm{m}}{{\mathrm{s}}^2}\right)t‐\left(3\times \frac{2\mathrm{m}}{{\mathrm{s}}^3}\right)t^2 \\ =\left(\frac{6.0\mathrm{m}}{{\mathrm{s}}^2}\right)t‐\left(\frac{6.0\mathrm{m}}{{\mathrm{s}}^3}\right)t^2 \end{gathered}$$

For$t=1.0\mathrm{s}$,

$$\begin{gathered} v\left(1.0s\right)=\left(6.0\times 1.0\right)-\left(6.0\times 1.0^2\right) \\ =0 \end{gathered}$$

Thus, the velocity at$t=1.0\mathrm{s}$ is $0\mathrm{m}/\mathrm{s}$.

Similarly,for$t=2.0\mathrm{s}$,

$$\begin{gathered} v\left(2.0s\right)=\left(6.0\times 2.0\right)-\left(6.0\times 2.0^2\right) \\ =-12\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the velocity at $t=2.0\mathrm{s}$is$-12\mathrm{m}/\mathrm{s}$

For $t=3.0\mathrm{s}$,

$$\begin{gathered} v\left(3.0s\right)=\left(6.0\times 3.0\right)-\left(6.0\times 3.0^2\right) \\ =-36\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the velocity at$t=3.0\mathrm{s}$ is $-36\mathrm{m}/\mathrm{s}$.

For$t=4.0\mathrm{s}$,

$$\begin{gathered} v\left(4.0s\right)=\left(6.0\times 4.0\right)-\left(6.0\times 4.0^2\right) \\ =-72\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the velocity at$t=4.0\mathrm{s}$ is$-72\mathrm{m}/\mathrm{s}$.

Equation of acceleration can be found by taking derivative of$v$with respect to$t$.

$$\begin{gathered} a=\frac{dv}{dt} \\ =\frac{d\left(2ct-3b{t}^2\right)}{dt} \\ =2c-6bt \\ =\left(2\times 3.0\right)-\left(6\times 2.0\right)t \\ =6.0-12t \end{gathered}$$

For$t=1.0\mathrm{s}$,

$$\begin{gathered} a\left(1.0s\right)=6.0-\left(12\times 1.0\right) \\ =-6.0\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the acceleration at $t=1.0\mathrm{s}$is$-6.0\mathrm{m}/{\mathrm{s}}^2$.

For $t=2.0\mathrm{s}$,

$$\begin{gathered} a\left(2.0s\right)=6.0-\left(12\times 2.0\right) \\ =-18\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the acceleration at $t=2.0\mathrm{s}$is $-18\mathrm{m}/{\mathrm{s}}^2$.

For$t=3.0\mathrm{s}$,

$$\begin{gathered} a\left(3.0s\right)=6.0-\left(12\times 3.0\right) \\ =-30\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the acceleration at $t=3.0\mathrm{s}$is $-30\mathrm{m}/{\mathrm{s}}^2$.

For$t=4.0\mathrm{s}$,

$$\begin{gathered} a\left(4.0\mathrm{s}\right)=6.0-\left(12\times 4.0\right) \\ =-42\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the acceleration at $t=4.0\mathrm{s}$is $-42\mathrm{m}/{\mathrm{s}}^2$.