Answer :
To determine the displacement of a particle [tex]\( P \)[/tex] when it is instantaneously at rest, we need to follow these steps:
1. Determine the velocity function:
The velocity of the particle is the first derivative of the displacement function [tex]\( x(t) \)[/tex] with respect to time [tex]\( t \)[/tex].
Given the displacement function:
[tex]\[ x(t) = \frac{1}{3} t(t^2 - 3t - 24) \][/tex]
Let's find the derivative of [tex]\( x(t) \)[/tex]:
[tex]\[ \frac{dx}{dt} = \frac{d}{dt} \left[ \frac{1}{3} t(t^2 - 3t - 24) \right] \][/tex]
Using the product rule, we get:
[tex]\[ \frac{dx}{dt} = \frac{1}{3} \left[ 1 \cdot (t^2 - 3t - 24) + t \cdot \frac{d}{dt}(t^2 - 3t - 24) \right] \][/tex]
[tex]\[ \frac{dx}{dt} = \frac{1}{3} \left[ (t^2 - 3t - 24) + t \cdot (2t - 3) \right] \][/tex]
[tex]\[ \frac{dx}{dt} = \frac{1}{3} \left[ t^2 - 3t - 24 + 2t^2 - 3t \right] \][/tex]
[tex]\[ \frac{dx}{dt} = \frac{1}{3} \left[ 3t^2 - 6t - 24 \right] \][/tex]
[tex]\[ \frac{dx}{dt} = t^2 - 2t - 8 \][/tex]
2. Find when the particle is at rest:
The particle is at rest when its velocity is zero. So, we set the velocity function equal to zero and solve for [tex]\( t \)[/tex]:
[tex]\[ t^2 - 2t - 8 = 0 \][/tex]
This is a quadratic equation. We can solve it using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -8 \)[/tex]:
[tex]\[ t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} \][/tex]
[tex]\[ t = \frac{2 \pm \sqrt{4 + 32}}{2} \][/tex]
[tex]\[ t = \frac{2 \pm \sqrt{36}}{2} \][/tex]
[tex]\[ t = \frac{2 \pm 6}{2} \][/tex]
This gives us two solutions:
[tex]\[ t = \frac{2 + 6}{2} = 4 \quad \text{and} \quad t = \frac{2 - 6}{2} = -2 \][/tex]
Since [tex]\( t \)[/tex] must be greater than or equal to zero ([tex]\( t \geq 0 \)[/tex]), we discard [tex]\( t = -2 \)[/tex]. Thus, [tex]\( t = 4 \)[/tex].
3. Determine the displacement when the particle is at rest:
Now, we plug [tex]\( t = 4 \)[/tex] back into the displacement function [tex]\( x(t) \)[/tex] to find the displacement at this time:
[tex]\[ x(4) = \frac{1}{3} \cdot 4 \left[(4)^2 - 3(4) - 24 \right] \][/tex]
[tex]\[ x(4) = \frac{1}{3} \cdot 4 \left[16 - 12 - 24 \right] \][/tex]
[tex]\[ x(4) = \frac{1}{3} \cdot 4 \cdot (-20) \][/tex]
[tex]\[ x(4) = \frac{1}{3} \cdot (-80) \][/tex]
[tex]\[ x(4) = - \frac{80}{3} \][/tex]
[tex]\[ x(4) = - \frac{80}{3} \approx -26.6667 \][/tex]
So, the displacement of particle [tex]\( P \)[/tex] when it is instantaneously at rest is [tex]\(- \frac{80}{3} \)[/tex] meters or approximately [tex]\( -26.6667 \)[/tex] meters.
1. Determine the velocity function:
The velocity of the particle is the first derivative of the displacement function [tex]\( x(t) \)[/tex] with respect to time [tex]\( t \)[/tex].
Given the displacement function:
[tex]\[ x(t) = \frac{1}{3} t(t^2 - 3t - 24) \][/tex]
Let's find the derivative of [tex]\( x(t) \)[/tex]:
[tex]\[ \frac{dx}{dt} = \frac{d}{dt} \left[ \frac{1}{3} t(t^2 - 3t - 24) \right] \][/tex]
Using the product rule, we get:
[tex]\[ \frac{dx}{dt} = \frac{1}{3} \left[ 1 \cdot (t^2 - 3t - 24) + t \cdot \frac{d}{dt}(t^2 - 3t - 24) \right] \][/tex]
[tex]\[ \frac{dx}{dt} = \frac{1}{3} \left[ (t^2 - 3t - 24) + t \cdot (2t - 3) \right] \][/tex]
[tex]\[ \frac{dx}{dt} = \frac{1}{3} \left[ t^2 - 3t - 24 + 2t^2 - 3t \right] \][/tex]
[tex]\[ \frac{dx}{dt} = \frac{1}{3} \left[ 3t^2 - 6t - 24 \right] \][/tex]
[tex]\[ \frac{dx}{dt} = t^2 - 2t - 8 \][/tex]
2. Find when the particle is at rest:
The particle is at rest when its velocity is zero. So, we set the velocity function equal to zero and solve for [tex]\( t \)[/tex]:
[tex]\[ t^2 - 2t - 8 = 0 \][/tex]
This is a quadratic equation. We can solve it using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -8 \)[/tex]:
[tex]\[ t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} \][/tex]
[tex]\[ t = \frac{2 \pm \sqrt{4 + 32}}{2} \][/tex]
[tex]\[ t = \frac{2 \pm \sqrt{36}}{2} \][/tex]
[tex]\[ t = \frac{2 \pm 6}{2} \][/tex]
This gives us two solutions:
[tex]\[ t = \frac{2 + 6}{2} = 4 \quad \text{and} \quad t = \frac{2 - 6}{2} = -2 \][/tex]
Since [tex]\( t \)[/tex] must be greater than or equal to zero ([tex]\( t \geq 0 \)[/tex]), we discard [tex]\( t = -2 \)[/tex]. Thus, [tex]\( t = 4 \)[/tex].
3. Determine the displacement when the particle is at rest:
Now, we plug [tex]\( t = 4 \)[/tex] back into the displacement function [tex]\( x(t) \)[/tex] to find the displacement at this time:
[tex]\[ x(4) = \frac{1}{3} \cdot 4 \left[(4)^2 - 3(4) - 24 \right] \][/tex]
[tex]\[ x(4) = \frac{1}{3} \cdot 4 \left[16 - 12 - 24 \right] \][/tex]
[tex]\[ x(4) = \frac{1}{3} \cdot 4 \cdot (-20) \][/tex]
[tex]\[ x(4) = \frac{1}{3} \cdot (-80) \][/tex]
[tex]\[ x(4) = - \frac{80}{3} \][/tex]
[tex]\[ x(4) = - \frac{80}{3} \approx -26.6667 \][/tex]
So, the displacement of particle [tex]\( P \)[/tex] when it is instantaneously at rest is [tex]\(- \frac{80}{3} \)[/tex] meters or approximately [tex]\( -26.6667 \)[/tex] meters.
