Question

5. The motion of a particle is defined by the relation x = 6t4 − 2t3 −12t2 + 3t + 3, where x and t are expressed in meters and seconds, respectively. Determine the time, the position, and the velocity when a = 0.

Answer 1

Distance X= 6t⁴ - 2t³ - 12t² +3t +3
Velocity V = dX/dt = 24t³ - 6t² - 24t +3
Acceleration a = dV/dt= 72t² - 12t - 24.
According to the question,
__72t² - 12t - 24 =0
or, 6t²-t - 2 =0
or, 6t² - 4t +3t - 2 =0
or, 2t(3t-2)+1(3t-2)=0
or, (2t+1)(3t-2) =0
Since time can't be negative so 3t-2 =0
The time required is = t= 2/3 seconds
Distance X = 6× (2/3)⁴ - 2(2/3)³ - 12(2/3)² +3(2/3) +3
= 32/27 - 16/27 - 16/3 + 6 + 3 = 9 - 16× 8/27 = 4.74 m
Velocity V= 24 × (2/3)³ - 6(2/3)² - 24(2/3) +3
_______V = 64/9 - 24/9 - 16 +3 = - 77/9 = - 8.56 m/sec
Hope this is ur required answer.

Answer 2

Given:

The relation $x=6t^{4}- 2t^{3}- 12t^{2}+ 3t+3$

To find:

The time, position and the velocity of the particle when acceleration is zero

Solution:

Step 1

We have been given an equational relation between the position and time of a particle in motion.

We know,

Velocity of a particle is the rate of change of position of the particle.

$v=\frac{dx}{dt}$  

Acceleration of the particle is the rate of change of velocity of the particle.

$a=\frac{dv}{dt}$   $;$ $or$

Acceleration is the second derivative of the displacement of the particle.

$a=\frac{d^{2}x }{dt^{2} }$

Step 2

Now,

According to the question, we have,

$x=6t^{4}- 2t^{3}- 12t^{2}+ 3t+3$

If we take the derivative of the displacement of the particle with respect to time, we get the velocity.

$v=\frac{d(6t^{4}- 2t^{3}- 12t^{2}+ 3t+3)}{dt}$

$v=24t^{3}- 6t^{2} -24t+3$                      $(i)$

Now, if we take the derivative of velocity with respect to time, we get the acceleration of the particle.

$a=\frac{d(24t^{3}- 6t^{2} -24t+3)}{dt}$

$a=72t^{2} -12t-24$                             $(ii)$

Step 3

According to the question, acceleration of the particle is 0. Hence, we get

$72t^{2} -12t-24=0$

$6t^{2} -t-2=0$

$6t^{2} +3t-4t-2=0$

$3t(2t+1)-2(2t+1)=0$

$(3t-2)(2t+1)=0$

$t=\frac{2}{3},\frac{-1}{2}$

Time cannot be negative. Hence, acceleration will be zero at time $t=\frac{2}{3}sec$.

Step 4

Now,

The position of the particle at $t=\frac{2}{3}sec$ is given by,

$x=6t^{4}- 2t^{3}- 12t^{2}+ 3t+3$

$x=6(\frac{2}{3})^{4}-2(\frac{2}{3})^{3}-12(\frac{2}{3})^{2} +3(\frac{2}{3})+3$

$x=6(\frac{16}{81} )-2(\frac{8}{27} )-12(\frac{4}{9} )+2+3$

$x=\frac{32}{27}- \frac{16}{27}-\frac{16}{3}+5$

$x=\frac{16}{27}-\frac{1}{3}$

$x=\frac{7}{27}$

$x=0.259m$

Hence, the position of the particle at $t=\frac{2}{3}sec$ is 0.259 meters.

Step 5

Now,

The velocity of the particle at $t=\frac{2}{3}sec$ is

$v=24t^{3}- 6t^{2} -24t+3$

$v=24(\frac{2}{3} )^{3}-6(\frac{2}{3}) ^{2} -24(\frac{2}{3})+3$

$v=24(\frac{8}{27} )-6(\frac{4}{9} )-\frac{48}{3}+3$

$v=\frac{64}{9}- \frac{24}{9}-16+3$

$v=\frac{40}{9}-13$

$v=-8.5m/s$

Hence, the velocity of the particle at $t=\frac{2}{3}sec$ will be 8.5 m/s. The negative sign indicates the direction of the motion of the particle.

Final answer:

Hence,

1. The time when acceleration is zero is $\frac{2}{3}$ seconds.
2. The particle is at 0.259 meters at $\frac{2}{3}$ seconds.
3. The velocity of the particle at $\frac{2}{3}$ seconds is 8.5 m/s.