# What is the acceleration of the particle

1. Dec 4, 2014

### Ashes Panigrahi

1. The problem statement, all variables and given/known data
The motion of a particle is given as follows :
$x=u(t-2)+a(t-2)^2$
Which of the following option(s) is/are correct ?
A.The initial velocity of particle is u
B.The acceleration of the particle is a
C.The acceleration of the particle is 2a
D.At $t=2s$ particle is at origin.

2. Relevant equations
$x=ut+\frac{1}{2}at^2$

3. The attempt at a solution
Simplifying the equation, we get,
$x=ut-2u+at^2-4at+4a$
Could somebody guide me as to what will be the next step ?

2. Dec 4, 2014

### Staff: Mentor

Have you had (or are studying) calculus?

Chet

3. Dec 5, 2014

### Ashes Panigrahi

No. I have had no exposure to calculus.

4. Dec 5, 2014

### lep11

Well, how are velocity and acceleration defined and related to each other?

5. Dec 5, 2014

### Ashes Panigrahi

The mentioned scenarios can be reduced to relative motion. Right ?

6. Dec 5, 2014

Nope.

7. Dec 5, 2014

### Staff: Mentor

Well, without calculus, it's not as easy , but try this: substitute t'=t-2 in your original equation and see shat you get.

Chet

8. Dec 5, 2014

### Ashes Panigrahi

But there is no t'. I just simplified the problem equation.

9. Dec 5, 2014

### Staff: Mentor

Let's try something different. Let's take your simplified equation and rewrite it as:
$$x=(4a-2u)+(u-4a)t+at^2$$
Compare this with:
$$x=x_0+v_0t+\frac{1}{2}At^2$$
where
x0=particle location at time zero
v0=particle velocity at time zero
A = particle acceleration

Chet

10. Dec 6, 2014

### lep11

dx/dt=v
dv/dt=a
I think you may be familiar with derivatives even if you haven't had much exposure to calculus.