What is the acceleration of the particle

[ubergewehr273]

Homework Statement

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.

Homework Equations

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

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 ?

 

[Chestermiller]

Have you had (or are studying) calculus?

Chet

 

[ubergewehr273]

No. I have had no exposure to calculus.

 

[lep11]

No. I have had no exposure to calculus.

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

 

[ubergewehr273]

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

 

[lep11]

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

Nope.

 

[Chestermiller]

No. I have had no exposure to calculus.

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

 

[ubergewehr273]

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

 

[Chestermiller]

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

 

[lep11]

Simplifying the equation, we get,
$x=ut-2u+at^2-4at+4a$

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