# Mechanics Question

• Mar 13th 2009, 09:38 AM
djmccabie
Mechanics Question
A particle of mass 3kg, moves along the horizontal x-axis under the action of a resultant force FN. Its velocity V m/s at time t s.

v=12t-3t²

a) given that the particle is at the origin O when t=1, find an expression for the displacement of the particle from O at the time t s.

b) Find the acceleration of the particle at time t s.

c) Find the power of the force F when t=1.5

right for part a I think i am supposed to integrate v so i do

∫12t-3t² dt (Is this correct? as in am i supposed to write dt? or dx?)

this gives me the result x=6t²-t³ where x = displacement

For part b do i differentiate the original equation?

No sure how to do part C

Any help welcome =]
• Mar 13th 2009, 09:49 AM
hairymclairy
Hey,

You're perfectly right about parts a and b, with a) you need to integrate with t going from 0 to 1.

For part c) you need to use The equation F=ma where F is the force, m is mass and a is acceleration.

Hope that helps.
• Mar 13th 2009, 10:00 AM
Quote:

Originally Posted by hairymclairy
Hey,

For part c) you need to use The equation F=ma where F is the force, m is mass and a is acceleration.

For Part C the Power is given by

P(t) =F(t).v(t)

Where p is power v is velocity and f is force at time t

Hope this adds to that help (Giggle)
• Mar 13th 2009, 10:03 AM
djmccabie
thanks for helping. So before i attempt part c, are the answers to parts a and b:

a) x=6t²-t

b) A= 12-6t

where you said t going from 0 to 1, are these the limits to put on the integral?

also would the answer to part c be 101.25W ?
• Mar 13th 2009, 10:25 AM
Quote:

Originally Posted by djmccabie
A particle of mass 3kg, moves along the horizontal x-axis under the action of a resultant force FN. Its velocity V m/s at time t s.

v=12t-3t²

a) given that the particle is at the origin O when t=1, find an expression for the displacement of the particle from O at the time t s.

b) Find the acceleration of the particle at time t s.

c) Find the power of the force F when t=1.5

right for part a I think i am supposed to integrate v so i do

∫12t-3t² dt (Is this correct? as in am i supposed to write dt? or dx?)

this gives me the result x=6t²-t³ where x = displacement

For part b do i differentiate the original equation?

No sure how to do part C

Any help welcome =]

Displacement = ∫12t-3t² dt

$
S(t) = 12(t^2)/2 - 3(t^3)/3 + c$

Now put the limit from 0 to 1

S(1) - S(0) = 6 - 1 = 5m

----------------------------

v=12t-3t²

$Acceleration = \frac{dv}{dt}$

$= 12 - 6t$

-----------------------------------

$Power = F(t).v(t)$

$=3a(t).v(t) = 3(12-6t)(12t-3t^2)$

Now put $t= 1.5s$
• Mar 13th 2009, 10:49 AM
djmccabie
Projectiles
thanks