Graphical Analysis of Motion: 1. Refer to the graphs to...

Graphical Analysis of Motion: 1. Refer to the graphs to answer the following questions. Vy (m/s) 3.5 1.0 0.0 1.3 t (s) Vy (m/s) 5.5 1.5 0.75 1.3 t (s) Calculate the displacement and acceleration for the motion described for each above graphs, respectively. 2. An object is tossed vertically and its motion is plotted on the following graph. Answer the following questions using the graph: a. Calculate the object's acceleration. b. At what time does the object reach its highest point? Vy (m/s) 4.5 -5.5 1.1 t (s) c. Calculate the displacement for the object when it reaches its highest point. d. Find the total displacement for the object from t=0s to 1.1s. Then sketch the corresponding position-time graph.

Transcript

00:01 Hello student we have given these two graphs and we have to find the displacement and acceleration under both these graph okay so for both the graphs displacement is equal to what displacement is equal to area under graph area under graph okay and what is the acceleration so acceleration this is equal to this is represented by a and displacement is represented by d this is equal to what? slope under graph okay slope of this is what acceleration is okay now we are considering graph a that is our a part so in graph a the displacement d will be equals to half multiplied by 1 .3 multiplied by 2 .5 okay plus 1 multiplied by 1 .3 so this is the area under the graph so this will be equals to what 2 .925 meter okay this is the displacement now the acceleration is the slope so slope is y2 minus y1 divided by x2 minus x1 so this will be course to 2 2 .5 okay y2 minus y1 is 2 .5 divided by 1 .3 so this is equals to 1 .923 meter per second square now we are considering the graph b okay in b part we have to find the same displacement okay so what is this displacement this is equals to half multiplied by 4 multiplied by 0 .55 plus 1 .5 multiplied by jr .5 so what is this equals to? so this equals to 1 .925 meter.

01:31 So this is the displacement for the b.

01:33 Now we will find the acceleration which is equals to y2 minus y1 divided by x2 minus x1 which is equals to 4 divided by 0 .55 and this value came as 7 .273 meter per second is so this is the answer for the b part now in the question two okay first part we have to calculate what acceleration so what is the acceleration equals to this is equals to slope of graph okay this is equals to what slope of graph and this is equals to minus 9 .5 divided by 1 .1 equals to minus 8 .636 36 meter per second square okay now what we have to find in the b part in b part we have to find the velocity right so we use the equation of line so what is the equation of line over there so this is equals to v mt plus c right where m is the source so this will be caused to minus 8 .636 multiplied by the time plus c is 4 .5.

02:31 Okay, so this is what the velocity, okay? but at time equals to 0, okay, at which the velocity is 0, we have to find that time...

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