Another particle moving in a straight line problem

In summary, the particle has a velocity of 8 m s-1, and a displacement of 32 cm from the origin when it passes through the point at time t=2.
  • #1
karush
Gold Member
MHB
3,269
5
A particle, moving in a straight line, passes through a fixed point O
with a velocity of $8 m s^{-1}$
Its acceleration, $a\ m^{-2}$, t seconds after passing O is given by $a=12-6t$. Find
(i) the velocity of the particle when $t=2$,
(ii) the displacement of the partial from O when $t=2$.

I know that if $a=12-6t$.
Then $v=12t+3t^2+C$
 
Last edited:
Physics news on Phys.org
  • #2
karush said:
A particle, moving in a straight line, passes through a fixed point O with a velocity of $8 m s^{-1}$
In acceleration, $a$ in $m\,\mathbf{s}^{-2}$, t seconds after passing O is given by $a=12-6t$. Find
(i) the velocity of the particle when $t=2$,
(ii) the displacement of the partial from O when $t=2$.

I know that if $a=12-6t$.
Then $v=12t+3t^2+C$

It would really help if you posted the question properly. I have fixed what I think you meant...

Anyway, since you know that when the point passes through the origin, the velocity is 8, that should help you work out your integration constant.
 
  • #3
karush said:
A particle, moving in a straight line, passes through a fixed point O
with a velocity of $8 m s^{-1}$
Its acceleration, $a\ m^{-2}$, t seconds after passing O is given by $a=12-6t$. Find
(i) the velocity of the particle when $t=2$,
(ii) the displacement of the partial from O when $t=2$.

I know that if $a=12-6t$.
Then $v=12t+3t^2+C$

You have been given the IVP:

\(\displaystyle a(t)=\d{v}{t}=12-6t\) where \(\displaystyle v_0=8\frac{\text{m}}{\text{s}}\)

Another approach, would be to use the initial and final conditions (boundaries) as the limits of integration:

\(\displaystyle \int_{v_0}^{v(t)}\,du=\int_{t_0}^{t}12-6w\,dw\)

Now, we can choose to set the time at which the particle passes point $O$ as time $t=0$ and write:

\(\displaystyle \int_{8}^{v(t)}\,du=\int_{0}^{t}12-6w\,dw\)

Integrating, we find:

\(\displaystyle v(t)-8=12t-3t^2\)

Or:

\(\displaystyle v(t)=-3t^2+12t+8\)

Can you put together the IVP to find the position (displacement) function $x(t)$?
 
  • #4
$$v\left(2\right)=20 \frac{m}{s}$$
Which is the answer to (i)

But how do we get displacement if we don't have $s=$
 
  • #5
karush said:
$$v\left(2\right)=20 \frac{m}{s}$$
Which is the answer to (i)

But how do we get displacement if we don't have $s=$

We could let the initial position be 0, that is:

\(\displaystyle x(0)=0\)

But we don't have to...we could simply say $x(0)=x_0$, and then the displacement would be:

\(\displaystyle \Delta x=x(t)-x_0\)
 
  • #6
Well then $s\left(t\right)=-{t}^{3}+6{t}^{2}+8t$
Since $C=0$
So $s\left(2\right)=32 $
 
  • #7
What I would do is write:

\(\displaystyle v(t)=\d{x}{t}=-3t^2+12t+8\)

Use the boundaries as the limits, and replace the dummy variables:

\(\displaystyle \int_0^{x(t)}\,du=\int_0^t -3w^2+12w+8\,dw\)

Utilize the FTOC:

\(\displaystyle x(t)=-t^3+6t^2+8t\)

And so:

\(\displaystyle x(2)=-8+24+16=32\)

And this agrees nicely with your result. :)
 

FAQ: Another particle moving in a straight line problem

What is a particle moving in a straight line problem?

A particle moving in a straight line problem is a physics problem that involves calculating the position, velocity, or acceleration of an object that is moving in a straight line. It is a fundamental concept in physics and is often used to model the motion of objects in real-world scenarios.

How do you calculate the position of a particle in a straight line problem?

The position of a particle can be calculated using the equation x = x0 + v0t + ½at2, where x0 is the initial position, v0 is the initial velocity, a is the acceleration, and t is the time.

What is the difference between velocity and speed in a particle moving in a straight line problem?

Velocity is a vector quantity that describes the rate of change of an object's position in a particular direction, while speed is a scalar quantity that describes the rate of change of an object's position without regard to direction. In a particle moving in a straight line problem, velocity is typically used to describe the motion of the particle.

How do you calculate the acceleration of a particle in a straight line problem?

The acceleration of a particle can be calculated using the equation a = (v - v0) / t, where v is the final velocity, v0 is the initial velocity, and t is the time. This equation assumes that the acceleration is constant throughout the motion of the particle.

What are some real-world examples of particle moving in a straight line problems?

Some common examples of particle moving in a straight line problems include the motion of a car on a straight road, the falling of an object due to gravity, and the launch of a rocket into space. These problems can be solved using the principles of physics and the equations of motion for a particle in a straight line.

Similar threads

Replies
4
Views
2K
Replies
2
Views
1K
Replies
3
Views
2K
Replies
8
Views
987
Replies
4
Views
1K
Replies
3
Views
2K
Replies
45
Views
5K
Back
Top