[SOLVED] Differential Equation

**cyt91** · February 22nd 2010, 01:34 AM

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!

**Grandad** · February 22nd 2010, 10:16 AM

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)