The position of a particle at time t sec is s = t^3-6t^2+9t-5 Find the displacement during the first 5 seconds, find the average velocity during the first 5 seconds and find the acceleration for each time the particles velocity is zero

Answer 1

To analyze the motion of the particle defined by the position function $s(t)=t^3-6t^2+9t-5$, we'll go through the steps you've provided.

Step 1: Find the displacement during the first 5 seconds.

The displacement during a time interval is given by the difference between the position at the end of the interval and the position at the beginning of the interval.

1. Calculate $s(0)$: $$s(0)=0^3-6(0^2)+9(0)-5=-5$$

2. Calculate $s(5)$: $$s(5)=5^3-6(5^2)+9(5)-5$$ $$=125-150+45-5=15$$

3. Calculate displacement: $$\text{Displacement}=s(5)-s(0)=15-(-5)=15+5=20$$

Step 2: Find the average velocity during the first 5 seconds.

The average velocity over a time interval is given by the total displacement divided by the length of the time interval.

$$\text{Average Velocity}=\frac{\text{Displacement}}{\mathrm{\Delta }t}=\frac{20}{5}=4,\text{m/s}$$

Step 3: Find the acceleration for each time the particle's velocity is zero.

First, we need to find the velocity function, which is given by the derivative of the position function: $$v(t)=\frac{ds}{dt}=\frac{d}{dt}(t^3-6t^2+9t-5)=3t^2-12t+9$$

To find when the velocity is zero, set $v(t)=0$: $$3t^2-12t+9=0$$ Dividing the equation by 3: $$t^2-4t+3=0$$

Factoring: $$(t-1)(t-3)=0$$ Thus, $$t=1\text{and}t=3$$

Next, we calculate the acceleration, which is the derivative of the velocity function, $a(t)=\frac{dv}{dt}=\frac{d}{dt}(3t^2-12t+9)$: $$a(t)=6t-12$$

Now, we can find the acceleration at the points where the velocity is zero:

1. For $t=1$: $$a(1)=6(1)-12=6-12=-6,{\text{m/s}}^2$$

2. For $t=3$: $$a(3)=6(3)-12=18-12=6,{\text{m/s}}^2$$

Summary of Results:

• Displacement during the first 5 seconds: 20 meters
• Average velocity during the first 5 seconds: 4 m/s
• Acceleration when velocity is zero:
  • At $t=1$: $-6,{\text{m/s}}^2$
  • At $t=3$: $6,{\text{m/s}}^2$