You are in front of the class, moving back and forth in a straight line to demon

Question

You are in front of the class, moving back and forth in a straight line to demonstrate a new, improved sonic ranger. The ranger is set up to measure positive position if you are to the right of the center of the room, and negative position if you are left of the center of the room. As you move, the device produces the following graph of velocity vs. time What can you tell about your position at the instant t = 1 sec? Why? What is your acceleration at t=3 sec? Be sure to state in your own words what does acceleration mean (not just write down the equation!). What is your total displacement between t=0 and t = 3 sec? What is your average velocity between t=2 and t = 3.5 sec?

Explanation / Answer

1) the object starts from centre and starts moving towards right with a constant velocity of 4m/sec for 2 seconds and then its velocity decreases to 0 in 1 second . after that it starts moving towards left and its velocity increases to 2m/sec towards left and then continues its journey at this velocity.

hence at t=4.5sec you are moving towards left.

2) from t=0 to t=1, v=4m/sec

now if it starts from origin then position at t=1 sec is 4x1m =4m.

3) acceleartion at t=3 sec is the slope of v-t graph at t=3 seconds . now slope= 6/5.5=1.09

this much will be the acceleration , a=1.09m/sec2

4) total displacement between t=0 and t = 3 sec= area under v-t curve for that duration

D =(2x4)+(4x1)/2=10m

5) average velocity between t=2 and t = 3.5 sec=(total displacement)/(total time taken)

=11/1.5 m/sec= 7.33m/sec

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