# Graphical Analysis

1. Jan 2, 2008

### rejz55

View attachment Doc1.doc

how am i going to interpret the distance-vs-time and the acceleration-vs-time of this graph?
Thanks!
(the y axis is the velocity and the x is the time- attached file)

2. Jan 2, 2008

### Shooting Star

The dist will be the usual area under the curve, with proper signs.

The accn is the slope of the curve, all constants in this case, with a few points where the accn is undefined.

3. Jan 2, 2008

### rejz55

Can you elaborate your explanation pls..Thanks..I have my few interpretations here..I just do no Know if they are right..Thanks..(see the attached file)

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4. Jan 3, 2008

### Shooting Star

In a speed vs. time graph, the area under the curve gives you the distance, since the distance is integral vdt. I hope you know this. The accn is dv/dt, and so the slope of the tangent at a point represents the accn.

Look at the start: the v is -5 m/s for some time t1 (not specified). So, the particle is moving left (say), and the dist covered will be the product of -5*t1 (in metres). It’s negative.

Next, at a single instant, the speed becomes +5 from -5, showing that accn is infinitely large and not defined at that instant. Typically, this represents an impulsive force acting for a very short time (what we call collisions).

Then the speed decreases linearly from 5 to 0, and the accn is uniform because the slope of the graph is constant. The accn a= (0-5)/t2 m/s/s. The dist is the area of the triangle, giving the formula d= ½ v0*t2.

Then the speed is zero => dist covered is 0 and accn =0.

You get the drift, I hope. At the end, you can sum up all the distances, with proper signs, to find the total dist travelled. Find the accns for each of the segments by finding the slope.

5. Jan 3, 2008

### rejz55

Thanks shooting star..