# Average Acceleration and Velocity Question

1. Jul 17, 2012

### CathyCat

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

An object traveling at a constant velocity v0 experiences a constant acceleration in the same direction for a period of time t. Then, an acceleration of equal magnitude is experienced in the opposite direction of v0 for the same period of time t. What is the object's final velocity? State the magnitude and direction.

2. Relevant equations

Acceleration=Vf-Vi/t

3. The attempt at a solution

i don't know what they are asking for, as in i dont understand the question
Thank you

Last edited: Jul 17, 2012
2. Jul 17, 2012

### CathyCat

can someone help me please. its a choice question as in the magnitude is a)more than initial magnitude, b)less than initial magnitude, and c)the same as the initial magnitude
and for the direction its a) same as initial magnitude or b) opposite of initial magnitude

3. Jul 17, 2012

### PeterO

To paraphrase the question:
You are travelling in a car on a straight road, heading East at 40 kph.
You press the accelerator so the car now accelerates at a rate of 5 kph/s East. You continue pressing the accelerator for 5 seconds [and will thus reach a new, higher speed.
You then carefully apply the brakes, so as to achieve an acceleration of 5 kph/s West [naturally the braking force/acceleration is in the direction opposed to the way you are travelling]. You maintain that braking rate for exactly 5 seconds.
What will your final velocity be? ie how fast will you be travelling, and are you going East or West?

If you are uncomfortable with kph [kilometres per hour] then change it to mph [miles per hour] but retain the figures - it just means a more significant speed at all times, but still not unreasonable.

Note:
While travelling East, an easterly acceleration means increasing speed [what usually happens when you press the accelerator].
While travelling East, a westerly acceleration means reducing speed or slowing down [what usually happens when you apply the brakes].