# Kinematics

• May 30th 2008, 10:23 AM
Simplicity
Kinematics
Question:
A particle moves along a straight line starts at time $\displaystyle t=0$ seconds with a velocity $\displaystyle 4 \ ms^{-1}$. At any subsequent time $\displaystyle t$ seconds the acceleration of the particle is $\displaystyle (6t-8) \ ms^{-2}$. Find:
(a) The distance moves before first coming to instantaneous rest,
(b) the total time $\displaystyle T$ seconds taken by the particle to return to the starting point,
(c) the greatest speed of the particle for $\displaystyle 0 \le t \le T$.

(a) $\displaystyle x = \frac{32}{27} \ m$
(b) $\displaystyle T = 2 \ s$

My Problem in this question:
I don't understand part (c). Here is my method:

$\displaystyle v = 3t^2 - 8t + 4$
$\displaystyle \therefore$ Greatest speed when $\displaystyle \frac{\mathrm{d}v}{\mathrm{d}t} = 0$. (At turning point?)
$\displaystyle \therefore \frac{\mathrm{d}v}{\mathrm{d}t} = 6t-8=0 \implies t = \frac43$
$\displaystyle \therefore v = 3\left(\frac43\right)^2 - 8\left(\frac43\right) + 4 = -\frac43$
$\displaystyle \text{Speed} = |v| = \frac43$ ?!!!

How would part (c) be done? The correct answer is $\displaystyle 4 \ ms^{-1}$. Thanks in advance.
• May 30th 2008, 10:39 AM
bobak
Quote:

Originally Posted by Air
Question:
A particle moves along a straight line starts at time $\displaystyle t=0$ seconds with a velocity $\displaystyle 4 \ ms^{-1}$. At any subsequent time $\displaystyle t$ seconds the acceleration of the particle is $\displaystyle (6t-8) \ ms^{-2}$. Find:
(a) The distance moves before first coming to instantaneous rest,
(b) the total time $\displaystyle T$ seconds taken by the particle to return to the starting point,
(c) the greatest speed of the particle for $\displaystyle 0 \le t \le T$.

(a) $\displaystyle x = \frac{32}{27} \ m$
(b) $\displaystyle T = 2 \ s$

My Problem in this question:
I don't understand part (c). Here is my method:

$\displaystyle v = 3t^2 - 8t + 4$
$\displaystyle \therefore$ Greatest speed when $\displaystyle \frac{\mathrm{d}v}{\mathrm{d}t} = 0$. (At turning point?)
$\displaystyle \therefore \frac{\mathrm{d}v}{\mathrm{d}t} = 6t-8=0 \implies t = \frac43$
$\displaystyle \therefore v = 3\left(\frac43\right)^2 - 8\left(\frac43\right) + 4 = -\frac43$
$\displaystyle \text{Speed} = |v| = \frac43$ ?!!!

How would part (c) be done? The correct answer is $\displaystyle 4 \ ms^{-1}$. Thanks in advance.

Your asked to find the maximum speed in the given interval 0-2 seconds.

as $\displaystyle v = 3t^2 - 8t + 4$ the maximum speed is unbounded with time increasing, your calculus method found the minimum speed of the particle. Draw a rough sketch of the velocity function and the answer will be obvious.

Bobak