# Thread: Particle Motion - Beginner

1. ## 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

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.

2. Just use Newton's Second Law.

3. Originally Posted by Ackbeet
Just use Newton's Second Law.
How? Thank you.

4. 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?

5. 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

6. 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!

7. F = M x a

Therefore a = F/M.

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

8. 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?)

9. m/s^2 ?

I hope lol

10. Looks good to me.

11. 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

x = 2 t^2 + t - sin t

12. 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.

13. 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:

(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.

14. 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.