1. ## Kinematics

A particle travelling in a straight line passes a fixed point O with a velocity of 8m/s and a accerlation of -2m/$\displaystyle s^2$. It moves in such a manner that t seconds after passing O its accerlation, fm/s^2 , is given by f= a + bt . During the 1st second after passing O it moves through 9m. calculate the value of a and of b.

2. Originally Posted by helloying
A particle travelling in a straight line passes a fixed point O with a velocity of 8m/s and a accerlation of -2m/$\displaystyle s^2$. It moves in such a manner that t seconds after passing O its accerlation, fm/s^2 , is given by f= a + bt . Mr F says: I like using a for acceleration so I'm gonna change this to $\displaystyle {\color{red}\alpha + \beta t}$.

During the 1st second after passing O it moves through 9m. calculate the value of a and of b. I dont really understand this question. This qn is so long and i cant imagine how the particle move.
Let $\displaystyle t = 0$ be when the particle passes the fixed point O. Then you have:

$\displaystyle a = \alpha + \beta t$ subject to the boundary conditions that when $\displaystyle t = 0 \, \text{s}$, $\displaystyle x = 0 \, \text{m}$, $\displaystyle v = 8 \, \text{m/s}$ and $\displaystyle a = -2 \, \text{m/s}^2$.

$\displaystyle t = 0 \, \text{s}$: $\displaystyle a = \alpha \Rightarrow \alpha = -2$. Therefore $\displaystyle a = -2 + \beta t$.

$\displaystyle a = \frac{dv}{dt} = -2 + \beta t \Rightarrow v = -2t + \frac{\beta}{2} t^2 + C$.

Substitute $\displaystyle v = 8 \, \text{m/s}$ when $\displaystyle t = 0 \, \text{s}$: $\displaystyle C = 8$. Therefore $\displaystyle v = -2t + \frac{\beta}{2} t^2 + 8$.

$\displaystyle v = \frac{dx}{dt} = -2t + \frac{\beta}{2} t^2 + 8 \Rightarrow x = -t^2 + \frac{\beta}{6} t^3 + 8t + K$.

Substitute $\displaystyle x = 0 \, \text{m}$ when $\displaystyle t = 0 \, \text{s}$: $\displaystyle K = 0$.

Therefore $\displaystyle x = -t^2 + \frac{\beta}{6} t^3 + 8t$.

Now substitute $\displaystyle x = 9$ when $\displaystyle t = 1$ and solve for $\displaystyle \beta$.