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/ $s^2$. 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/ $s^2$. 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 ${\color{red}\alpha + \beta t}$.

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 $t = 0$ be when the particle passes the fixed point O. Then you have:

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

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

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

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

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

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

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

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