Q15P - Fundamentals Of Physics

Question

(a) If a particle’s position is given by $\mathbf{x}\mathbf{=}\mathbf{4}\mathbf{-}\mathbf{12}\mathbf{t}\mathbf{+}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}$ (where t is in sec and x in metres), what is its velocity at $\mathbf{t}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{1}\mathbf{ }\mathbf{sec}$ ? (b) Is it moving in positive or negative direction of x just then? (c) What is its speed just then? (d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculations) (e) Is there ever an instant when the velocity is zero? If so, give the time t, if no, answer no. (f) Is there a time after $\mathbf{t}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{3}\mathbf{ }\mathbf{sec}$ when the particle is moving in negative direction of x? If so, give the time t, if no, answer no.

Step-by-Step Solution

Answer

(a) Velocity of particle at $\mathrm{t}=1\mathrm{s}$  is $-6 \mathrm{m}/\mathrm{s}$ 

(b) Direction of motion of particle is in negative x direction.

(c) Speed of particle at   $\mathrm{t}=1\mathrm{s}$ is $6 \mathrm{m}/\mathrm{s}$.

(d) Speed of particle at $\mathrm{t}=0$ to $2 \mathrm{s}$ is decreasing and at $\mathrm{t}=3 \mathrm{s}$  it is increasing.

(e) An instant velocity of particle at $\mathrm{t}=2 \mathrm{s}$ is 0 .

(f) No. when  $\mathrm{t} >0$then the velocity is $\mathrm{v}>0$ .

Step 1: Given data

The position of the particle is given by the equation, 

$\mathrm{x}=4-12\mathrm{t}+3{\mathrm{t}}^2$

Step 2: Understanding the relationship between displacement and velocity

The velocity can be found by differentiating the displacement with respect to time. Once the equation is derived for the velocity, substitute the values of time and displacement to find the value. The equation for the velocity in terms of displacement is,

$\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}$

Step 3: (a) Calculate the velocity at t = 1   sec

Take the derivative of x  to find velocity to get,

$$\begin{gathered} \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}} \\ =\frac{\mathrm{d}}{\mathrm{dt}}\left(4-12\mathrm{t}+3{\mathrm{t}}^2\right) \\ =-12+6\left(\mathrm{t}\right) \end{gathered}$$

 

Therefore,

  $\mathrm{v}= -12+6\left(\mathrm{t}\right) \left(\mathrm{i}\right)$                                       

Substitute the value of t to find velocity.

$\mathrm{v}=- 6\hspace{0.33em}\mathrm{m}/\mathrm{s}$ 

Therefore, the velocity of the particle at $\mathrm{t}=1\mathrm{s}$ is $-6 \mathrm{m}/\mathrm{s}$.

Step 4: (b) Finding the direction of motion of the particle.

From the step 3, velocity is less than 0, $\mathrm{v}<0$, the particle is moving along -x  direction at $\mathrm{t}=1 \mathrm{s}$ .

Step 5: (c) Calculating the speed at t = 1 &#160; s e c

At given time $\mathrm{t}=1 \mathrm{s}$, the velocity is $-6 \mathrm{m}/\mathrm{s}$ . Therefore, the magnitude of velocity, i.e. speed is $\left|\mathrm{v}\right|=6 \mathrm{m}/\mathrm{s}$.

Step 6: (d) Find out if the speed is increasing or decreasing

From the equation $\mathrm{v} = -12+6\mathrm{t}$, we see that at $\mathrm{t} = 2\hspace{0.33em}\mathrm{s}$  velocity becomes 0 .

So, from $0\hspace{0.33em}\mathrm{s} \mathrm{to} 2\hspace{0.33em}\mathrm{s}$ velocity decreasing and at $\mathrm{t} = 3\hspace{0.33em}\mathrm{s}$ we get velocity 6 m/s that means it is increasing.

Step 7: (e) Find out the instant at which velocity becomes zero

The equation for the velocity is given as,

 $\mathrm{v}= -12+6\left(\mathrm{t}\right)$

 

Substitute the velocity equals to zero.

 $$\begin{gathered} 0=-12+6\left(t\right) \\ \mathrm{t} = 2\hspace{0.33em}\mathrm{s} \end{gathered}$$

Therefore, at $\mathrm{t} = 2\hspace{0.33em}\mathrm{s}$  velocity becomes 0 .

Step 8: (f) Finding the time after &#160; t = 3 &#160; sec &#160; when the particle is moving in the negative direction of x

Using equation (i), it can be seen that for  $\mathrm{t} >2\hspace{0.33em}\mathrm{s} , \mathrm{v} > 0.$, 

Therefore, after  3 sec, velocity will never be zero.