Interpreting Distance-vs-Time and Accel-vs-Time Graphs

[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)

 

[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.

 

[rejz55]

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.

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)

 

[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.

 

[rejz55]

Thanks shooting star..