# Motion in one dimension problem

1. Nov 1, 2012

### Ammar w

1. The problem statement, all variables and given/known data
An object moves along the x axis according to the equation x = 3.00t2 - 2.00t + 3.00 , where x is in meters and t is in seconds.
Determine :
(a) the average speed between t = 2.00 s and t = 3.00 s.
(b) the instantaneous speed at t = 2.00
(c) the average acceleration between t = 2.00 s and t = 3.00 s.
(d) the instantaneous acceleration at t = 2.00 s
(e) at what time is the object at rest?

2. Relevant equations
average speed = $\frac{d(total distance travelled)}{\Delta t}$

instantaneous speed = $\frac{dx}{dt}$

$\overline{a}$ (average acceleration) = $\frac{\Delta v}{\Delta t}$

a (instantaneous acceleration) = $\frac{dv}{dt}$ = $\frac{dx^2}{dt}$

3. The attempt at a solution

(a) I couldn't calculate the total distance.

(b) instantaneous speed = instantaneous velocity = $\frac{dx}{dt}$ = 6.00t - 2.00
instantaneous speed at t=2.00 = 6.00(2)-2.00 = 10 m/s

(c) $\overline{a}$ = $\frac{\Delta v}{\Delta t}$ = $\frac{vf-vi}{1}$ = $\frac{v(3)-v(2)}{1}$ = $\frac{16-10}{1}$ = 6 m/s2

(d) a = $\frac{dx^2}{dt}$ = 6 => a at 2.00s = 6 m/s2

(e) the rest means that v = 0
=> 6t-2=0
=>t=0.33 s //it seems wrong.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 1, 2012

### AJ Bentley

Looks like you did pretty well.

The total distance? x= distance,
so calculate x at 3 secs minus x at 2 secs

3. Nov 1, 2012

### Ammar w

thanks
but x at 3 minus x at 2 is the displacement which we use to calculate the average velocity not average speed. right??

4. Nov 1, 2012

### AJ Bentley

Speed is just (absolute) velocity.

5. Nov 1, 2012

### Ammar w

I think no.\
the average velocity is different from average speed
average velocity = dispacement/time
average speed = distance/time
sometimes average velocity is zero (when a particle moves and return to the starting point) where the average speed is a positive number.

in this problem I think the distance is like what you said = x at 3s - x at 2s because the position x is always increasing (from the equation)

6. Nov 1, 2012

### AJ Bentley

Speed is defined as the scalar magnitude of the velocity vector. Since this is a one dimensional problem it simply becomes the absolute velocity.
Fact.

7. Nov 1, 2012

### Ammar w

sorry, still not convinced!