1. ## Kinematics (Calculus Form)

Question:
A particle travels along a straight line which passes through $A$ and $B$. During the motion the velocity of the particle is $6t+t^2$ metres per second, where $t$ seconds is the time measured from a certain instant. The particle passes through $A$ when $t=2$ and through $B$ when $t=5$. Find, in terms of $t$, the acceleration of the particle and its distance from $A$ at any instant during the motion. Calculate, also, the distance of $B$ from $A$.

$a = \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}(6t+t^2)}{\mathrm{d}t} = 6 + 2t$

$v = (6t+t^2) = \frac{\mathrm{d}x}{\mathrm{d}t}$
$\therefore x = 3t^2 + \frac13 t^3 + c$
$x_{\vec{AB}} = \left[3t^2 + \frac13 t^3\right]^5_2 = 102$

My Problem:
I don't know how to do this part (don't even understand what it is saying ): 'Find...its distance from $A$ at any instant during the motion'*. Can someone help? Thanks in advance.

*The answer is $x = 3t^2+ \frac13 t^3 - \frac{44}{3}$.

2. Originally Posted by Air
Question:
A particle travels along a straight line which passes through $A$ and $B$. During the motion the velocity of the particle is $6t+t^2$ metres per second, where $t$ seconds is the time measured from a certain instant. The particle passes through $A$ when $t=2$ and through $B$ when $t=5$. Find, in terms of $t$, the acceleration of the particle and its distance from $A$ at any instant during the motion. Calculate, also, the distance of $B$ from $A$.

$a = \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}(6t+t^2)}{\mathrm{d}t} = 6 + 2t$

$v = (6t+t^2) = \frac{\mathrm{d}x}{\mathrm{d}t}$
$\therefore x = 3t^2 + \frac13 t^3 + c$
$x_{\vec{AB}} = \left[3t^2 + \frac13 t^3\right]^5_2 = 102$

My Problem:
I don't know how to do this part (don't even understand what it is saying ): 'Find...its distance from $A$ at any instant during the motion'*. Can someone help? Thanks in advance.

*The answer is $x = 3t^2+ \frac13 t^3 - \frac{44}{3}$.
The particle is at A when t = 2. So
$x_A = 3(2)^2 + \frac13 (2)^3 + c = \frac{44}{3} + c$

Thus the particle's distance from A will be
$x - x_A = \left ( 3t^2 + \frac13 t^3 + c \right ) - \left ( \frac{44}{3} + c \right ) = 3t^2 + \frac{1}{3}t^3 - \frac{44}{3}$

-Dan

3. Originally Posted by topsquark
The particle is at A when t = 2. So
$x_A = 3(2)^2 + \frac13 (2)^3 + c = \frac{44}{3} + c$

Thus the particle's distance from A will be
$x - x_A = \left ( 3t^2 + \frac13 t^3 + c \right ) - \left ( \frac{44}{3} + c \right ) = 3t^2 + \frac{1}{3}t^3 - \frac{44}{3}$

-Dan
How did you know that it is $x - x_A$ and not $x_A - x$?

4. Originally Posted by Air
How did you know that it is $x - x_A$ and not $x_A - x$?
When $t > 2, \, x > x_A$ and so clearly the distance is $x - x_A$. HOWEVER .......

When $0 < t < 2, \, x < x_A$ and so the distance will be $x_A - x$ ...... It's apparent that the question actually wanted the distance from at any instant during the motion for t > 2.