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 - 3

*t* , 0 <=

*t* <= 1

-(4

*x*+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 $\displaystyle 0\le t \le 1$, we have:$\displaystyle a = \frac{dv}{dt}=5-3t$

For $\displaystyle 0\le t \le 1, a > 0$. So the particle's speed is increasing during this phase.

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

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

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

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

$\displaystyle \Rightarrow v(1) = 5 -\tfrac32 = \tfrac72$

Then, writing $\displaystyle v$ as $\displaystyle \frac{dx}{dt}$:$\displaystyle \frac{dx}{dt}=5t-\tfrac32t^2$

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

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

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

$\displaystyle \Rightarrow x(1) = 2$

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

(b)

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

$\displaystyle \Rightarrow \tfrac12v^2 = -2x^2-x + c$

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

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

$\displaystyle \Rightarrow \tfrac12v^2 = -2x^2-x+\frac{129}{8}$

When $\displaystyle x = 2.5$:$\displaystyle \tfrac12v^2 = -\frac{25}{2}-\frac{5}{2}+\frac{129}{8}$$\displaystyle =\frac98$

So when $\displaystyle x = 2.5$ the speed is $\displaystyle \sqrt{\frac94}=1.5$ cm/sec.

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

Grandad