Question

The position of a particle moving along x-axis is related to time / as follow: x = $2t^2-t^3$, where x is in meters and t is in seconds.
a.What is the maximum positive displacement of the particle along the x axis and at what instant does it attain it?
b.Describe the motion of the particle.
c.What is the distance covered in the first three seconds?
d.What is its displacement in the first four seconds?
e.What are the particles average speed and average velocity in the first 3 seconds?
f.What are the particles instantaneous acceleration at the instant of its maximum positive x displacement? 1
g.What is the average acceleration between the interval
t = 2s to t = 4s?

Answer 1

The derivative of x w.r.L time will give the velocity of the particle in relation with time.

$v=\frac{dx}{dt}=4t-3t^2$ ...(i)

The double derivative of x w.r.t time will give acceleration of the particle in relation with time

$a=\frac{d^{2}x}{d{t}^2}=4-6t$ ...(ii)

a. For minimum and maximum displacement, dx/dt = 0.

Thus,

$\frac{dx}{dt}=4t-3t^2=0$

t(4 - 3t) = 0 $\Rightarrow$ either t = 0 or $t=\frac{4}{3}s$

Also, for maxima, double derivative of x should be negative, i.e., $\frac{d^{2}x}{d{t}^2}<0$

$\frac{d^{2}x}{d{t}^2}=4-6t$

At t = 0; $\frac{d^{2}x}{d{t}^2}$ = 4 - 6(0) = 4 (positive)

And at t = (4/3)s; $\frac{d^{2}x}{d{t}^2}$ = 4 - 6 ($\frac{4}{3}$) = 4 - 8 = -4 (negative)

Therefore,the particle is at maximum displacement at t =$\frac{4}{3}$ s and the corresponding displacement is

$x_{max}$ = 2${\left(\frac{4}{3}\right)}^2$ - ${\left(\frac{4}{3}\right)}^3$ = $\frac{32}{27}m$

Note: The maximum particle displacement of the particle is $x_{max}$ = $\frac{32}{27}m$

and it will occur at t = $\frac{4}{3}s$.

b.Velocity is positive when v > 0 or 4t - 3$t^2$ > 0 or t(4 - 3t) > 0

i.e.,t > 0 and t <$\frac{4}{3}s$

For acceleration to be positive, a > 0 $\Rightarrow$ 4 - 6t >0 or t < $\frac{2}{3}s$

From t = 0 to t < $\frac{2}{3}s$, the velocity ($\overrightarrow{v}$) and ($\overrightarrow{a}$) both are in

the same direction. Hence, velocity of the particle increases continuously during the time interval t = 0 to t = $\frac{2}{3}s$.

From t = $\frac{2}{3}s$ onward, the acceleration ($\overrightarrow{a}$) acts in negative x-direction. Hence, the velocity of the particle decreases 4

and becomes zero at t = $\frac{4}{3}s$. Then after, the particle moves in negative x-direction.

c. The velocity is positive between the instants 0 and 4/3 s and is negative at instant 4/3 s to the remaining 3 s.

Total distance = $\left|Displacementfrom0to\frac{4}{3}s\right|$ + $\left|Displacementfrom0to\frac{4}{3}sto3s\right|$

i.e., $|x\left(\frac{4}{3}\right)-x(0\left)\right|+\left|x\right(3)-x\left(\frac{4}{3}\right)|$

=$|2{\left(\frac{4}{3}\right)}^2-{\left(\frac{4}{3}\right)}^3-(0\left)\right|$ + $|2(3{)}^2-(3{)}^2-\{2{\left(\frac{4}{3}\right)}^2-{\left(\frac{4}{3}\right)}^3\}|$

=$\frac{16}{9}(2-\frac{4}{3})+\left|9\right(2-3)-\frac{16}{9}(2-\frac{4}{3})|=\frac{307}{27}m$

d. The required displacement is difference in position at t = 4s and t = 0.