# Particle Motion - Beginner

• May 23rd 2011, 05:55 AM
LukeEvans
Particle Motion - Beginner
Hi, I'm currently revising and I have a question relating to Particle Motion. The question grows, but to get started I have to show that

Quote:

The acceleration, a m/s^2, of the particle, is

a = sin t + 4

Given that m = 6 (kg) and F(t) = 6 sin t + 24N

Words do not express how grateful for any help I would be! I'd like to understand how to do this. Thanks.
• May 23rd 2011, 06:22 AM
Ackbeet
Just use Newton's Second Law.
• May 23rd 2011, 06:32 AM
LukeEvans
Quote:

Originally Posted by Ackbeet
Just use Newton's Second Law.

How? Thank you.
• May 23rd 2011, 06:35 AM
Ackbeet
Well, you've got F(t) = 6 sin t + 24 [N]. Newton's Second Law says that F = m a. Therefore, you can solve for a and plug in your F and m, and the result pops right out. Can you show the steps?
• May 23rd 2011, 06:40 AM
LukeEvans
Quote:

Originally Posted by Ackbeet
Well, you've got F(t) = 6 sin t + 24 [N]. Newton's Second Law says that F = m a. Therefore, you can solve for a and plug in your F and m, and the result pops right out. Can you show the steps?

so I get

6 sin t + 24 [N] = 6 x a

a = 6 sin t + 24[N] / 6

at t = 0, a = 0.406

• May 23rd 2011, 06:47 AM
Ackbeet
Well, all you've shown is the value of the acceleration at t = 0. That doesn't finished the job, at least as stated in the OP. Incidentally, you really need to put parentheses in your expression. What you've actually written is

a = 6 sin t + (24[N] / 6),

when you really meant

a = (6 sin t + 24[N]) / 6.

To finish the problem, just divide the 6 out term-by-term. You're almost there!
• May 23rd 2011, 06:51 AM
LukeEvans
F = M x a

Therefore a = F/M.

(6Sint + 24N)/6 = Sin t + 4
• May 23rd 2011, 06:54 AM
Ackbeet
Quote:

Originally Posted by LukeEvans
F = M x a

Therefore a = F/M.

(6Sint + 24N)/6 = Sin t + 4

There you go. Don't forget to tag on the correct units at the end (which are what?)
• May 23rd 2011, 06:57 AM
LukeEvans
m/s^2 ?

I hope lol
• May 23rd 2011, 07:01 AM
Ackbeet
Looks good to me.
• May 23rd 2011, 07:09 AM
LukeEvans
Marvelous, thanked accordingly good sir!

Unsurprisingly, I have a further question; assuming the particle starts from rest at any datum point, how do I show that the distance traveled, (x), from the datum point at time t is

Quote:

x = 2 t^2 + t - sin t
• May 23rd 2011, 08:18 AM
TheEmptySet
Quote:

Originally Posted by LukeEvans
Marvelous, thanked accordingly good sir!

Unsurprisingly, I have a further question; assuming the particle starts from rest at any datum point, how do I show that the distance traveled, (x), from the datum point at time t is

Here is a hint: you know the acceleration, but acceleration is the derivative of velocity so this gives

$\frac{dv}{dt}=\sin(t)+4$

So now take the integral to get the velocity and don't forget that v(0)=0(the particle started from rest.)

Now do the same thing to find the position! See if you can finish from here.
• May 23rd 2011, 08:37 AM
LukeEvans
Quote:

Originally Posted by TheEmptySet
Here is a hint: you know the acceleration, but acceleration is the derivative of velocity so this gives

$\frac{dv}{dt}=\sin(t)+4$

So now take the integral to get the velocity and don't forget that v(0)=0(the particle started from rest.)

Now do the same thing to find the position! See if you can finish from here.

Forgive me for being confused. How does that look? I have broken the question down, this is it in full:

Quote:

(b) Given that initially the particle starts from rest at a particular datum point on the line of motion, show that the distance travelled, x, from the datum point by the particle at time t is,

x = 2 t^2 + t -sin t

Hence, show that the time taken for the particle to travel 8 m from the datum point is given by the solution of the equation f(t)=0 where:

f(t) = 2 t^2 + t - 8 -sin t
I just need to see it happen step by step. Thanks again.
• May 23rd 2011, 10:21 AM
TheEmptySet
Quote:

Originally Posted by LukeEvans
Forgive me for being confused. How does that look? I have broken the question down, this is it in full:

I just need to see it happen step by step. Thanks again.

You just need to integrate both sides of the equation with respect to t

$\int \frac{dv}{dt}dt=\int (\sin(t)+4)dt \implies v(t)=-\cos(t)+4t+c$

Now use the initial condition

$v(0)=0$

to solve for c.

Now just do the exact same thing except this time velocity is the derivative of position this gives

$\frac{dx}{dt}=$

The answer you got above. See if you can finish from here.