| A Velocity vs Time graph lends itself to both a quantitative and qualitative
analysis. What information can be determined by analyzing the graph below |
| which is the same graph drawn by the animator? |
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| Qualitatively, one can look at the graph and determine how fast the
object is moving at any given time (instantaneous velocity). You can also
determine |
| the direction in which the object is moving. In the graph above, from
0 - 2.00 s, the object is moving vertically as in the case of a ball thrown
straight up |
| with a uniform decrease in speed. The object is moving in a positive
direction because the velocity is positive although it is slowing down.
At |
| 2.00 s, the object stops for an instant before reversing direction.
From 2.00 - 4.00 s, the object is moving with a negative velocity while
the object is |
| moving faster and faster. This graph could also show an object moving
to the right (away from the origin in a positive direction), stopping
for an |
| instant to reverse direction, and from 2.00 - 4.00 s it is moving towards
the origin in a negative direction. |
| |
| Qualitatively, one can get an idea of the magnitude of an object's displacement.
The area between the graph and the time axis equals an object's |
| displacement. The area shaded blue above the x-axis will be positive
because both the velocity and the time are greater than zero. The area
shaded blue |
| below the x-axis will be negative because the velocity is negative and
the time is greater than zero. It is easy to see that these two shaded
areas are equal |
| in magnitude and vectorially add to give a zero displacement. |
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| |
| Quantitatively, one can determine the magnitude of an object's displacement.
The two most common geometric shapes that you will encounter will be the |
| triangle and rectangle. |
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| To determine the displacement from 0 to 2.00 s, you would determine
the area of a triangle: |
| |
A = 1/2•b•h = 1/2•v•t = 1/2•19.60m/s•2.0
s = 19.6 m |
| |
| To determine the displacement from 2.00 to 4.00 s, you would determine
the area of a triangle: |
| |
A = 1/2•b•h = 1/2•v•t = 1/2•-19.60m/s•2.0
s = -19.6 m |
| |
| The quantitative result Δs = +19.6 m - 19.6 m = 0 agrees with
the qualitative result. |
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| If you are familiar with calculus, an alternative approach is to integrate
over a time interval. Because the area under a Velocity vs Time graph
equates to |
| the displacement, the definite integral of a velocity function gives
the displacement given by: |
| |
| Δs = ∫ v(t)dt where
the upper and lower limits would be t2 and t1 respectively. |
| |
| Mathematically, one can determine the instantaneous acceleration by
finding the slope of a Velocity vs Time graph. For example, what is the
average |
| acceleration at 3.0 s? Because a straight line has only one slope, the
average acceleration between 2.0 and 3.0 s will be equal to the instantaneous
|
| acceleration at 3.0 s: |
| |
| a = Δv/Δt = (-9.80 m/s - 0)/1.0 s = -9.80
m/s2 |
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| and the object is speeding up in the negative direction. |
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| When a graph is curved, it is a little more difficult to determine the
instantaneous velocity. At a time t, you must draw a tangent to the graph
making sure |
| there is only one point of intersection and determining the slope of
the tangent line. If you are familiar with calculus, it is even easier
since: |
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| a(t) = dv/dt |
| |
| because the derivative of a velocity function gives the acceleration. |
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