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

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

$a = 2t-6$

$\frac{dv}{dt} = 2t-6$

Integrate acceleration equation

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

$v = 2(\frac{t^2}{2})-6t+C_1$

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

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

$v = t^2-6t$

$\frac{dx}{dt} =t^2-6t$

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

$x = \frac{t^3}{3}-6\frac{t^2}{2}+C_2$

At t=0, x=0

$0 = \frac{0^3}{3}-6(\frac{0^2}{2})+C_2$

Therefore the value of the second constant is

$C_2 = 0$

$x = \frac{t^3}{3}-\frac{6t^2}{2}$

Calculate the distance traveled after 11 s

At  $t=11s$

$x = \frac{11^3}{3}-6(\frac{11^2}{2})$

$x = 80.667m$