Q18P - Fundamentals Of Physics

Question

The position of a particle moving along an x axis is given by $\mathbf{x}\mathbf{=}\mathbf{12}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{2}{\mathbf{t}}^{\mathbf{3}}$, where x in metres and t in sec. Determine (a) the position. (b) the velocity, and (c) the acceleration of the particle at t=3 s . (d) What is the maximum positive coordinate reached by the particle and (e) at what time is it reached? (f) What is the maximum positive velocity reached by the particle, and (g) at what time is it reached? (h) What is the acceleration of the particle at the instant the particle is not moving? (Other than t=0)? (i) Determine the average velocity of the particle between t=0 and t=3 s . 

Step-by-Step Solution

Answer

(a) The position at t=3.0 s is 54 m 

(b) Velocity at t=3.0 s is 18 m/s 

(c) Acceleration at t=3.0 s  is $-12\mathrm{m}/{\mathrm{s}}^2$

(d) Maximum positive coordinate is 64 m 

(e) Time at maximum positive coordinate is 4 s 

(f) Maximum positive velocity is 24 m/s 

(g) Time at maximum velocity is 2.0 s 

(h) Acceleration at maximum t is $-24\mathrm{m}/{\mathrm{s}}^2$ 

(i) Average velocity between t=0 and t=3 s  is 18 m/s

Step 1: Given information

$x\left(t\right)=12t^2-2t^3$

Step 2: To understand the concept of average velocity

The problem deals with the average velocity which is the displacement over the time. Here position of particle at any time t can be found by plugging this time in given equation. We can take derivative of the given position equation with respect to t to get equation of velocity and acceleration. Using condition v=0, we can find time at that point and then using that time we can find maximum coordinate. For maximum velocity, acceleration should be zero. With this, the time from acceleration equation and ultimately maximum velocity can be found.

 

Formula:

 

The velocity in general is given by,

$v=\frac{dx}{dt}$ 

The average velocity is given by,

${\mathrm{v}}_{\mathrm{avg}}=\frac{∆\mathrm{x}}{∆\mathrm{t}}$ 

Step 3: Derive expression for velocity and acceleration

$x\left(t\right)=12t^2-2t^3$

$\begin{array}{l}\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\\ =24\mathrm{t}-6{\mathrm{t}}^2\end{array}$

$\begin{array}{l}\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}\\ =24-12\mathrm{t}\end{array}$

Step 4: Calculate the position, velocity and acceleration at t=3 s

$x\left(t\right)=12t^2-2t^3$

By plugging the values we get position,

$$\begin{gathered} \mathrm{x}\left(\mathrm{t}\right)=12{\left(3.0\right)}^2-2{\left(3.0\right)}^3 \\ =54 \mathrm{m} \end{gathered}$$ 

 

Similarly,

Velocity at t=3.0 s,

$$\begin{gathered} \mathrm{v}\left(3\right)=24\left(3\right)-6{\left(3\right)}^2 \\ =18 \mathrm{m}/\mathrm{s} \end{gathered}$$ 

Acceleration at t=3.0 s,

$$\begin{gathered} \mathrm{a}\left(3\right)=24-12\left(3\right) \\ =-12 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$ 

Step 5: Calculate the maximum positive coordinate reached by the particle and the time at which it is reached

At maximum coordinate $v=0$.

By solving $\nu =0$  we get,

$0=24\mathrm{t}-6{\mathrm{t}}^2$ 

$\begin{array}{l}t=24I6\\ =4s\end{array}$ 

For this time, maximum position can be calculated by plugging the value in equation of position:

$x=12(4{)}^2-2(4{)}^3$ 

$\mathrm{x} = 64 \mathrm{m}$ 

Step 6: Calculate the maximum positive velocity reached by the particle and the time at which it is reached

For maximum positive velocity acceleration should be zero.

By solving acceleration equation for t we get,

$$\begin{gathered} t=24/12 \\ =2.0 \mathrm{s} \end{gathered}$$ 

 

Using this time in velocity equation we get,

$v=24 m/s$ 

Step 7: Calculate the acceleration of the particle at the instant the particle is not moving

At $t=4 s$ particle is motionless.

Using this time we get acceleration,

$$\begin{gathered} \mathrm{a}=24-12\left(4\right) \\ =-24\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$ 

Step 8: Calculate the average velocity of the particle between t=0 and t=3 s .

Using equation of average velocity to find velocity at t=0s to 3s 

$$\begin{gathered} \mathrm{v}=\frac{∆\mathrm{x}}{∆\mathrm{t}} \\ =\frac{\left(54-0\right)}{\left(3-0\right)} \\ =18 \mathrm{m}/\mathrm{s} \end{gathered}$$