Q22P - Fundamentals Of Physics

Question

The position of a particle moving along x axis depends on the time according to the equation $\mathit{x}\mathbf{=}\mathit{c}{\mathit{t}}^{\mathbf{2}}\mathbf{‐}\mathit{b}{\mathit{t}}^{\mathbf{3}}$, where x is in meters and t in seconds. What are the units of –(a)Constant c and (b)Constant b? Let their numerical values be 2.0 and 3.0 respectively. (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) $\mathit{t}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{s}$ .

Step-by-Step Solution

Answer

(a) Unit of constant c is $\mathrm{m}/{\mathrm{s}}^2$

(b) Unit of constant b is $\mathrm{m}/{\mathrm{s}}^3$.

(c) Time when particle reach maximum positive x position when c is 3.0 and b=2.0 is 1.0 s .

(d) The distance particle travels from $t=0.0 s$to 4.0 is 82 m .

(e) The displacement of particle when particle travels from $t=0.0 s$ to 4.0 is $-80 \mathrm{m}$ .

(f) The velocity of particle at $t=1.0 \mathrm{s} \mathrm{is} 0 \mathrm{m}/\mathrm{s}$ .

(g) The velocity of particle at $t=2.0 \mathrm{s} \mathrm{is} -12 \mathrm{m}/\mathrm{s}$ .

(h) The velocity of particle at $t=3.0 \mathrm{s} \mathrm{is} -36 \mathrm{m}/\mathrm{s}$ .

(i) The velocity of particle at $t=4.0 \mathrm{s} \mathrm{is} -72 \mathrm{m}/\mathrm{s}$ .

(j) Particle’s acceleration at $t= 1.0 \mathrm{s} \mathrm{is} -6 \mathrm{m}/{\mathrm{s}}^2$.

(k) Particle’s acceleration at $t=2.0 \mathrm{s} \mathrm{is} -18 \mathrm{m}/{\mathrm{s}}^2$.

(l) Particle’s acceleration at $t=3.0 \mathrm{s} \mathrm{is} -30 \mathrm{m}/{\mathrm{s}}^2$.

(m) Particle’s acceleration at $t=4.0 \mathrm{s} \mathrm{is} -42 \mathrm{m}/{\mathrm{s}}^2$ .

Step1: Approach to solving questionshould be as follow

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)

Step2: (a) Determination of the unit of constant c

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

Step 3: (b) Determination of the unit of b

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

Step 4: (c) Determination of the time when particle reaches to maximum x position

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

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

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

Step 5: (d) Determination of the distance traveled between the given time interval

For $t=4 s$ 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 $1.0 \mathrm{m}+1.0 \mathrm{m}+80 \mathrm{m}=82 \mathrm{m}$.

Step 6: (e) Determination of the displacement between the given time interval

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 $-80 \mathrm{m}$.

Step 7: (f) Determinationof velocity of particle at given time instant

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 $t=1.0 \mathrm{s}$,

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

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

Step 8: (g) Determinationof velocity of particle at given time instant

Similarly,for $t=2.0 \mathrm{s}$,

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

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

Step 9: (h) Determination of velocity of particle at given time instant

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

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

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

Step 10: (i) Determination of velocity of particle at given time instant

For $t=4.0 \mathrm{s}$,

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

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

Step 11: (j) Determination of acceleration of particle at given time instant

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 $t=1.0 \mathrm{s}$,

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

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

Step 12: (k) Determination of acceleration of particle at given time instant

For $t=2.0 \mathrm{s}$ ,

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

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

Step 13: (l) Determination of acceleration of particle at given time instant

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 $-30 \mathrm{m}/{\mathrm{s}}^2$.

Step 14: (m) Determination of acceleration of particle at given time instant

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 $-42 \mathrm{m}/{\mathrm{s}}^2$ .