## Velocity vs Time Primer

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? 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. 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. 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 mTo 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 mThe quantitative result Δs = +19.6 m - 19.6 m = 0 agrees with the qualitative result. 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 t _{2}and t_{1}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/s ^{2 }and the object is speeding up in the negative direction. 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: a(t) = dv/dt because the derivative of a velocity function gives the acceleration.