Question

Starting from rest, a particle moving in a straight line has an acceleration of a = (2t - 6) m/s^2, where t is in seconds. What is the particle's velocity when t = 6 s, and what is its position when t = 11 s?

Answer 1

The velocity of the particle at t = 6 s is 0 m/s, and its position at t = 11 s is 80.67 m.

Step-by-step explanation:

To find the velocity of the particle at t = 6 s, we need to integrate the acceleration function a(t) = (2t - 6) m/s². The acceleration can be integrated to yield the velocity function as follows:
V(t) = ∫ a(t) dt = ∫ (2t - 6) dt = t² - 6t + C

Since the particle starts from rest, we know that the initial velocity V(0) = 0, which allows us to solve for the constant C. Therefore, the velocity function is V(t) = t² - 6t.

The particle's velocity at t = 6 s is:

V(6) = 6² - 6(6) = 36 - 36 = 0 m/s

To determine the particle's position at t = 11 s, we must integrate the velocity function. This gives us the position function X(t):
X(t) = ∫ V(t) dt = ∫ (t² - 6t) dt = (1/3)t³ - 3t² + D

Since the particle starts from the origin, X(0) = 0, the constant D is 0. The position function then simplifies to X(t) = (1/3)t³ - 3t².

The particle's position at t = 11 s is:

X(11) = (1/3)(11)³ - 3(11)² = (1/3)(1331) - 3(121) = 443.67 - 363 = 80.67 m

Answer 2

To solve this problem we will use the given expression and derive it in order to find the algebraic expressions of velocity and position. These equations will be similar to those already known in the cinematic movement but will be subject to the previously given function. We start deriving the equation for velocity

`a = 2t-6`

`(dv)/(dt) = 2t-6`

Integrate acceleration equation

`\int dv = (2t-6)dt`

`v = 2((t^2)/(2))-6t+C_1`

`v=t^2-6t+C_1`

At
`t = 0, v = 0`

Replacing,

`0 = 0^2-6*0+C_1`

Therefore the value of the first Constant is

`C_1 = 0`

The expression can be escribed as,

`v = t^2-6t`

Calculate the velocity after 6s,

`v=t^2-6t`

`v = 6^2-6*6`

`v = 0m/s`

Now using the same expression we can derive the equation for distance

`v = t^2-6t`

`(dx)/(dt) =t^2-6t`

`\int dx = \int (t^2-6t)dt`

`x = (t^3)/(3)-6(t^2)/(2)+C_2`

At t=0, x=0

`0 = (0^3)/(3)-6((0^2)/(2))+C_2`

Therefore the value of the second constant is

`C_2 = 0`

`x = (t^3)/(3)-(6t^2)/(2)`

Calculate the distance traveled after 11 s

At
`t=11s`

`x = (11^3)/(3)-6((11^2)/(2))`

`x = 80.667m`