# kinematics differentiation (distance travelled by particle)

• May 13th 2011, 06:09 AM
arccos
kinematics differentiation (distance travelled by particle)
and so i got thrown into the sea without knowing how to swim again... (Speechless)

A particle moves in a straight line so that at time t seconds after leaving a fixed point of origin O, it's displacement is s metres and is governed by the equation $\displaystyle s = 8\cos(\dfrac{\pi}{3}t + \dfrac{\pi}{6})$. The initial velocity(v) of the particle is $\displaystyle -\dfrac{4(\pi)}{3} m/s$. Find the total distance travelled when the particle comes to rest at t = 2.5s

I kind of got near $\displaystyle 8 + 4\sqrt3$ which is the answer when i differentiated s twice to get the acceleration so i figured the answer is derived from there but i can't seem to get to the answer.. any pointers? thanks!
• May 13th 2011, 06:19 AM
alexmahone
Quote:

Originally Posted by arccos
and so i got thrown into the sea without knowing how to swim again... (Speechless)

A particle moves in a straight line so that at time t seconds after leaving a fixed point of origin O, it's displacement is s metres and is governed by the equation $\displaystyle s = 8\cos(\dfrac{\pi}{3}t + \dfrac{\pi}{6})$. The initial velocity(v) of the particle is $\displaystyle -\dfrac{4(\pi)}{3} m/s$. Find the total distance travelled when the particle comes to rest at t = 2.5s

I kind of got near $\displaystyle 8 + 4\sqrt3$ which is the answer when i differentiated s twice to get the acceleration so i figured the answer is derived from there but i can't seem to get to the answer.. any pointers? thanks!

$\displaystyle d=\int_{t=0}^{2.5}|v|dt$
• May 13th 2011, 06:45 AM
arccos
Quote:

Originally Posted by alexmahone
$\displaystyle d=\int_{t=0}^{2.5}|v|dt$

It worked :) But just wondering is there a way to solve it purely through differentiation? I got this question from the differentiation part of my book's kinematics exercises.
• May 13th 2011, 06:47 AM
Ackbeet
Quote:

Originally Posted by arccos
It worked :) But just wondering is there a way to solve it purely through differentiation? I got this question from the differentiation part of my book's kinematics exercises.

Just find the turning points, and find the distances from one turning point to the next. You can use the derivative to find turning points (the velocity will be zero there).
• May 13th 2011, 06:55 AM
arccos
Quote:

Originally Posted by Ackbeet
Just find the turning points, and find the distances from one turning point to the next. You can use the derivative to find turning points (the velocity will be zero there).

Thanks I will try that :)
To digress a little, I have a question where I have to determine the formula in t for displacement(s) given only the formula for velocity(v), how do I solve for C in the integral? No other information was given. ( v = -3t^2 + 8t + 5)
• May 13th 2011, 07:45 AM
Soroban
Hello, arccos!

This problem is much simpler than you think . . .

Quote:

A particle moves in a straight line so that at time $\displaystyle t$ seconds
after leaving a fixed point $\displaystyle O$, its displacement is $\displaystyle s$ metres
and is governed by the equation: .$\displaystyle s \:=\: 8\cos(\tfrac{\pi}{3}t + \tfrac{\pi}{6})$

The initial velocity of the particle is: .$\displaystyle v \:=\:-\tfrac{4\pi}{3}\text{ m/s}$
. . This is true, but irrelevant.

Find the total distance travelled when the particle comes to rest at $\displaystyle \,t = 2.5\:s$

Where is the particle at $\displaystyle t = 0$ ?

. . $\displaystyle s(0) \:=\:8\cos(\tfrac{\pi}{6}) \:=\:8(\tfrac{\sqrt{3}}{2}) \:=\:4\sqrt{3}$

Where is the particle at $\displaystyle t = \tfrac{5}{2}$ ?

. . $\displaystyle s(\tfrac{5}{2}) \:=\:8\cos\left[\tfrac{\pi}{3}(\tfrac{5}{2}) + \tfrac{\pi}{6}] \;=\; 8\cos\pi \;=\;8(\text{-}1) \;=\;\text{-}8$

$\displaystyle \text{The particle started at point }A\!: s = +4\sqrt{3}$
. . $\displaystyle \text{and stopped at point }B\!: s = \text{-}8.$
$\displaystyle \text{It }seems\text{ to have travelled a distance of: }\:8 + 4\sqrt{3}\text{ meters.}$

$\displaystyle \text{But did the particle go }directly\text{ from }A\text{ to }B\,?$
$\displaystyle \text{It is possible that the particle left point}\,A\text{, moved }beyond\text{ point}\,B,$
. . $\displaystyle \text{stopped, reversed direction and }returned\text{ to point}\,B.$
This, of course, would produce a greater total distance.

If the particle stopped and reversed direction.
. . then the velocity is zero at that time.

$\displaystyle v(t) \;=\;s'(t) \;=\;-\tfrac{8\pi}{3}\sin(\tfrac{\pi}{3}t + \tfrac{\pi}{6}) \:=\:0 \quad\Rightarrow\quad \sin(\tfrac{\pi}{3}t + \tfrac{\pi}{6}) \;=\;0$

. . $\displaystyle \tfrac{\pi}{3}t + \tfrac{\pi}{6} \:=\:\pi n \quad\Rightarrow\quad \tfrac{\pi}{3}t \:=\:\text{-}\tfrac{\pi}{6} + \pi n \quad\Rightarrow\quad t \:=\:\tfrac{3}{\pi}(\text{-}\tfrac{\pi}{6} + \pi n)$

. . $\displaystyle t \:=\:-\tfrac{1}{2} + 3n$

$\displaystyle \text{The first time the particle stops is when }n = 1$
. . $\displaystyle \text{that is: }\:t \:=\:\text{-}\tfrac{1}{2} + 3(1) \:=\:\tfrac{5}{2} \:=\:2.5\text{ sec}$

$\displaystyle \text{Therefore, the particle moved }directly\text{ from }A\text{ to }B.$
. . $\displaystyle \text{The total distance is: }\:8 + 4\sqrt{3}$