# Acceleration Problem

1. Jul 17, 2012

### CathyCat

1. The problem statement, all variables and given/known data

car accelerates from rest at a constant rate of 2.0 m/s^2 for 6.6 s.
(a) What is the speed of the car at the end of that time?
m/s

(b) How far does the car travel in this time?
m

2. Relevant equations

Acceleration=change in speed/time

3. The attempt at a solution
I did 2m/s^2=speed/6.6second for part a

and for part b, and just multiple 6.6 times 2 and got13.2

2. Jul 17, 2012

### PhanthomJay

3. Jul 17, 2012

### QED Andrew

$2 = \dfrac{v}{6.6}$ (where $v$ is the speed of the car at the end of that time) is the correct equation to find the answer to part (a) but isn't the answer to part (a). Do you see the difference?

To find the answer to part (b), the correct constant acceleration equation is $s = v_0 t + \frac{1}{2}at^2$ where $s$ is the displacement of car (what you were asked to find), $v_0$ is the initial speed of the car, $a$ is the acceleration of the car, and $t$ is the elapsed time.

The phrase "at rest" implies $v_0 = 0$, reducing the equation to $s = \frac{1}{2}at^2$.

From here, it is a matter of correctly evaluating the equation for the appropriate values of $a$ and $t$.

As PhanthomJay suggested, you should learn the constant acceleration equations (including and especially how they are derived).

4. Jul 18, 2012

### PeterO

Just wondering if you truly understand the m/s2 unit of acceleration.

The s2 part of the unit can be confusing - especially since it looks like the two s units for seconds is tied together.

Imagine you are the passenger in a car and you carefully watch the speedometer, and note at what rate your speed is building.

You may notice that the speed increases by 5 miles per hour each second - reaching 30 miles per hour in 6 seconds - not uncommon in city traffic.

That represents an acceleration of 5 (miles per hour) per second.
That could be written M/h/s [we wouldn't want to use m for miles as others may think you mean metres]

If the speedometer of the car was calibrated in metres per second [m/s] rather than miles per hour, you may have instead noticed the acceleration was 2 m/s each second, reaching 12 m/s in 6 seconds. [note that 12m/s is not all that different to 30 MPH; just a little slower]

That acceleration is just 2 (m/s)/s.
I used the brackets to remind you that the m and first s are connected in the unit of speed.
Without the brackets, that is written as m/s/s; which following standard mathematical techniques is written as m/s2 [at which point the meaning of the unit of acceleration becomes clouded, unfortunately]

Note: that last line should probably be best written m.s-1.s-1 which is "transcribed" using index notation to m.s-2