# Math Help - [SOLVED] Differential Equation

1. ## [SOLVED] Differential Equation

A particle moves along the +ve x-axis. At time t seconds after leaving a fixed point O from rest, the displacement of the particle from O is x cm.

The acceleration, a cm s^(-2) of the particle is defined by

a = { 5 - 3t , 0 <= t <= 1
-(4x+1), t>1

Find
(a) the maximum velocity of the particle and its distance from O when this happens

(b) the speed of the particle when x = 2.5

The acceleration of the particle is defined as a piece-wise function.

Forgive me for the rather unclear mathematical expressions. I'm new here and still learning the ropes of LATEX.

Btw, I calculated the maximum velocity to be 0.5 cm per second at distance 0.25 cm from O.
But the answer given is 3.5 cm per second at distance 2 cm from O.

Need help for both (a) and (b) !
Thank you!

2. Hello cyt91
Originally Posted by cyt91
A particle moves along the +ve x-axis. At time t seconds after leaving a fixed point O from rest, the displacement of the particle from O is x cm.

The acceleration, a cm s^(-2) of the particle is defined by

a = { 5 - 3t , 0 <= t <= 1
-(4x+1), t>1

Find
(a) the maximum velocity of the particle and its distance from O when this happens

(b) the speed of the particle when x = 2.5

The acceleration of the particle is defined as a piece-wise function.

Forgive me for the rather unclear mathematical expressions. I'm new here and still learning the ropes of LATEX.

Btw, I calculated the maximum velocity to be 0.5 cm per second at distance 0.25 cm from O.
But the answer given is 3.5 cm per second at distance 2 cm from O.

Need help for both (a) and (b) !
Thank you!
(a) For $0\le t \le 1$, we have:
$a = \frac{dv}{dt}=5-3t$
For $0\le t \le 1, a > 0$. So the particle's speed is increasing during this phase.

Then:
$\frac{dv}{dt}=5-3t$

$\Rightarrow v = 5t-\tfrac32t^2 + c$

$v(0) = 0 \Rightarrow c = 0$

$\Rightarrow v = 5t-\tfrac32t^2$

$\Rightarrow v(1) = 5 -\tfrac32 = \tfrac72$
Then, writing $v$ as $\frac{dx}{dt}$:
$\frac{dx}{dt}=5t-\tfrac32t^2$

$\Rightarrow x = \tfrac52t^2 -\tfrac12t^3 + c$

$x(0) = 0 \Rightarrow c = 0$

$\Rightarrow x = \tfrac52t^2 -\tfrac12t^3$

$\Rightarrow x(1) = 2$
Then for $t > 1$ and $x>0$, $a = -(4x+1) <0$. Hence the maximum positive velocity occurs at when $t = 1$; i.e. the maximum velocity is $3.5$ cm/sec when $x = 2$ cm.

(b)
$a = v\frac{dv}{dx}= -(4x+1)$

$\Rightarrow \tfrac12v^2 = -2x^2-x + c$
When $x = 2,\; v = \tfrac72$
$\Rightarrow \frac{49}{8}= -8-2+c$

$\Rightarrow c = \frac{129}{8}$

$\Rightarrow \tfrac12v^2 = -2x^2-x+\frac{129}{8}$
When $x = 2.5$:
$\tfrac12v^2 = -\frac{25}{2}-\frac{5}{2}+\frac{129}{8}$
$=\frac98$
So when $x = 2.5$ the speed is $\sqrt{\frac94}=1.5$ cm/sec.

Does that agree with the answer you've been given?