# Graph Problem on mechanics

1. Jun 20, 2009

### Mandavi

1. The problem statement, all variables and given/known data
Q-Figure shows the acceleration-time graph of a particle moving along a straight line.After what time the particle acquires its initial velocity?The graph is attached.

2. Relevant equations

We know that change in velocity is equal to the area covered by an acceleration-time graph.
ie (v-u)=at
3. The attempt at a solution
My logic is that the total change in velocity in 2 sec is 3 because (2+1/2*1*2).
So,if again initial velocity is to be reached,then the total change in velocity in negative direction should be 3.However,how should i calculate the time taken?

#### Attached Files:

• ###### graph2.bmp
File size:
286.7 KB
Views:
58
2. Jun 20, 2009

### Astronuc

Staff Emeritus
The velocity increases with contant acceleration. Then the velocity increases with linearly decreasing acceleration, and at some point the acceleration is actually deceleration (negative acceleration).

See - http://hyperphysics.phy-astr.gsu.edu/hbase/acons.html#c3

http://hyperphysics.phy-astr.gsu.edu/hbase/avari.html#c1

One needs to write the expression for acceleration for each period, and apply the appropriate limits to the integrals.

The period of constant acceleration is straightforward.

Velocity and displacement when acceleration is constant - http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html

3. Jun 20, 2009

### method_man

You can calculate like it's explained above and you can also calculate it by calculating area.
You said it yourself that the total change in velocity in negative direction should be 3.
If you have a triangle with a and b (take a look at picture I attached), the area under that triangle is 1/2*a*b=3. Compare that triangle with the one on the left in positive area to extract a (time).

#### Attached Files:

• ###### graph2.JPG
File size:
5.7 KB
Views:
60
4. Jun 22, 2009

### Mandavi

It was just a matter of applying similarity.Thanks a lot method_man!!:-)