Chapter 11, Problem 11.182RP

The motion of a particle is defined by the relation $x=2t^3-15t^2+24t+4$ , where x and t are expressed in meters and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and the total distance traveled when the acceleration is zero.

To determine

(a)

The time when the velocity is zero at given condition.

Answer to Problem 11.182RP

The required value of the time is $1s$ and $4s.$

Explanation of Solution

Given Information:

$x=2t^3-15t^2+24t+4$

Formula used:

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

$a=\frac{dv}{dt}$

Calculation:

We have given, $x=2t^3-15t^2+24t+4.$

Differentiating the above equation,

$\Rightarrow v=\frac{dx}{dt}=6t^2-30t+24$

Again differentiating the above equation:

$\Rightarrow a=\frac{dv}{dt}=12t-30$

Now when $v=0,$

$\Rightarrow 0=6t^2-30t+24=6\left(t^2-5t+4\right)$ ;

$\Rightarrow \left(t-4\right)\left(t-1\right)=0$ ;

$\Rightarrow t=1s$$t=4s.$

To determine

(b)

The position and the total distance traveled at given condition.

Answer to Problem 11.182RP

The required value of the position is $1.50m$ and the distance is $\mathrm{24.5.}$

Explanation of Solution

Given information:

$x=2t^3-15t^2+24t+4$

Formula used:

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

$a=\frac{dv}{dt}$

Calculation:

In order to find the position and distance traveled at $a=0.$

$\Rightarrow a=12t-30=0$$t=2.5s$ ;

$\Rightarrow x_2=2{\left(2.5\right)}^3-15{\left(2.5\right)}^2+24\left(2.5\right)+4$

Now the final position,

$\Rightarrow x=1.50m$

For, $0\le t\le 1s,v>0$

And for, $1s\le t\le 2.5s,v\le 0$

At, $t=0,x_0=4m$

$t=1s,x_1=\left(2\right){\left(1\right)}^3-\left(15\right){\left(1\right)}^2+\left(24\right)\left(1\right)+4=15m$

The distance traveled over interval is,

$\Rightarrow x_1-x_0=11m$

For, $1s\le t\le 2.5s$$v\le 0$

And the distance traveled over interval

$\Rightarrow \left|x_2-x_1\right|=\left|1.5-15\right|=13.5m$

Finally the total distance would be,

$\Rightarrow d=11+13.5$

$\Rightarrow d=24.5$.