Velocity Vs Time Analysis

Velocity vs Time Analysis

A Velocity vs Time graph lends itself to both a quantitative and qualitative analysis. What information can be determined by analyzing the graph below?

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, you could build a case for an object which 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 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 = ½bh = ½vt = ½*19.60m/s*2.00s= 19.6 m

To determine the displacement from 2.00 to 4.00 s, you would determine the area of a triangle:

A = ½bh = ½vt = ½*-19.60m/s*2.00s= -19.6 m

The 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 t2and t1respectively.

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

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

and the derivative of a velocity function gives the acceleration.

Velocity vs Time Analysis Problems

1) Use the graph below to answer the following questions.

(a) What is the displacement in segment 2?

To determine the displacement between 20.00 min and 50.00 min, you must find

the area of a rectangle, A = bh.

s = 150.00 m/min*30.00min= 4500. m

(b) What is the acceleration in segment 1?

To determine the acceleration, you find the slope of the line, m = Δy/Δx.

a = Δv/Δt = 150.00 m/min/20.00 min = 7.500 m/min2

There is no need to do a calculation so you look for a portion of the graph that

slopes down to the right which is segment 3.

2) Use the data and graph below to answer the following questions. The time between

each x-value is 0.50 s and the initial velocity is 20.0 m/s.

x (m)

0.00

9.12

16.50

22.12

26.00

28.12

(a) What is the average velocity during the second time interval?

One method of solution is to use the definition of average velocity:

vave= Δs/Δt = (16.50 m-9.12 m)/(1.00 s-0.50 s) = 14.76 m/s

Alternatively, you can find the acceleration to determine the instantaneous

velocities at points 2 and 3.

Δs = viΔt + ½aΔt2

9.12 m= 20.0 m/s*0.50s+ ½a(0.50s)2

a = -7.0 m/s2

vf= vi+ aΔt

vf= 20.0 m/s-7.0 m/s2*0.50s= 16.5 m/s

vf= 20.0 m/s-7.0 m/s2*1.00s= 13.0 m/s

vave= (vi+ vf)/2

vave= (16.5 m/s + 13.0 m/s)/2 = 14.75 s

(b) What is the acceleration?

See calculation above.

vf2= vi2+ 2aΔs

vf= ((20.0 m/s)2+ 2*-7.0 m/s2*28.12 m)1/2= 2.51 m/s

3) Use the graphs below to answer the following questions.

(1)(2)

(3)(4)

(a) For which graph(s) is the instantaneous velocity equal to the average velocity?

Graphs 1, 2, and 3 represent a constant velocity because the slope of each graph

is constant. When the velocity is constant, the instantaneous and average

velocities are always equal.

(b) Which graph(s) represent(s) a non-uniform velocity?

The slope of graph 4 is constantly increasing indicating a non-uniform velocity.

Graph 2 because the slope always equals zero.

(d) Which graph(s) represent(s) constant velocity?

Graphs 1, 2, and 3 for the reason given in Question 1.

4) During its road test, a Road Racer accelerates from rest to 30.0 m/s in 10.0 s. It then

travels at a constant velocity for an additional 10.0 s. For the last 10.0 s, the Road

Racer decelerates uniformly until it stops.

(a) Construct a Velocity vs Time graph for the Road Racer.

(b) Compare the total distance and the total displacement for the Road Racer. Justify

your answers.

Area of triangle 1:

Δs1= ½bh = ½*30.0 m/s*10.0s= 150. m

Area of rectangle 2:

Δs2= bh = 30.0 m/s*10.0s= 300. m

Area of triangle 3:

Δs3= ½bh = ½*30.0 m/s*10.0s= 150. m

ΔsT= 600. m

Both the distance and displacement are 600. m because according to the graph,

the car never reversed direction.

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