The figure shows the y-position (in blue) of a particle versus time.
(a) What is the average velocity of the particle during the time interval
t = 1.50 s
to
t = 4.50 s?
(Express your answer in vector form.)
Incorrect: Your answer is incorrect.
webMathematica generated answer key m/s
(b) Using the tangent to the curve (shown as the orange line in the figure), what is the instantaneous velocity of the particle at
t = 1.50 s?
(Express your answer in vector form.)
m/s
(c) At what time is the velocity of the particle equal to zero?
s
2 answers
These questions cannot be answered with the diagram that they are based on.
*For a) the point for 1.50s is (1.50s, 11m) and for 4.50s is (4.50s,17.5m)
*Plug in thee average velocity formula: Vav= delta y over delta t which is(17.5m-11m)/(4.50s-1.50s)=2.17jm/s
*For b) you need another point on the orange line, let chose point(3s,20m)
* Plug in the instaneous formula: S= delta y over delta t which is (20m-11m)/(3s-1.50s)=6.00m/s
*For c) Look at the highest point on the graph. In this case it is 3.50s.
*Plug in thee average velocity formula: Vav= delta y over delta t which is(17.5m-11m)/(4.50s-1.50s)=2.17jm/s
*For b) you need another point on the orange line, let chose point(3s,20m)
* Plug in the instaneous formula: S= delta y over delta t which is (20m-11m)/(3s-1.50s)=6.00m/s
*For c) Look at the highest point on the graph. In this case it is 3.50s.
1 answer