Acceleration Vs Time Analysis

Acceleration vs Time Analysis

An Acceleration vs Time graph lends itself to a more restricted analysis than does a Displacement vs Time or a Velocity vs Time graph. What information can be determined by analyzing the graph below?

Qualitatively, one can look at the graph and determine the rate of change of acceleration or deceleration (instantaneous acceleration). This graph shows a constant deceleration for the entire 4.0 s interval. As the projectile is moving upward in a positive direction, it is constantly decreasing in velocity at the rate of -9.80 m/s2. During its downward flight, the projectile is increasing in velocity but in a negative direction.

The area between the graph and the time axis equals the object's change in velocity, Δv. The area above the x-axis will be positive because a > 0 and t > 0. The area below the x-axis will be negative because a < 0 and t > 0.

To determine the Δv from 0 to 4.00 s, you would determine the area of a rectangle:

A = Δv = bh = aΔt = -9.80 m/s2*4.0s= -39.2 m/s

This is easy to remember starting with the kinematics formula: vf= vi+ aΔt

vf- vi= aΔt

Δv = aΔt

Don't overlook the y-intercept because it tells you the initial conditions, when the observer started to collect data. In the graph above, att = 0, a = g = -9.80 m/s2. Keep in mind that the initial value of t does not have to equal zero.

Acceleration vs Time Analysis Problems

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

(1)(2)

(3)(4)

(a) Which graph(s) show the object moving to the right at a constant velocity?

Graphs 1, 2, and 4 because they have a positive slope, i.e. the graph slopes up to

theright. Graphs 2 and 4 indicate the objectsstart from the origin (reference

point) with 4 having the greater velocity. Graph 1 indicates the object starts at the

left of the origin and moves to the right with a uniform velocity.

(b) Which graph(s) show the object moving with a constant positive acceleration?

No graphs show a uniform positive acceleration because all the slopes are

constant. If the slope of a D vs T graph is constant, then the velocity must be

constant. A D vs T graph exhibits uniform acceleration when the shape of the

graph is parabolic.

Movingto the right at a constant velocity?

Graph 1 because it is the only graph with a negative displacement.

(d) Which graph(s) show the object starting to the right of the reference point and

moving to the right at a constant acceleration?

None, see the answer in (b).

2) Use the graph below to answer the following questions. Give a reason to support

your answers.

(a) Which object has the highest velocity?

Object 3 has the highest velocity because it has the greatest slope.

(b) Are objects 1, 2, and 3 ever at the same point at the same time?

No because the three graphs do not intersect at a common point.

Object 3 is approximately 4.2 m ahead; 14.2 m - 10.00 m.

(d) How far did object 3 travel between the times it passed 2 and 1?

Object 3 travelsapproximately 2.8 m, (9.00 m - 6.20 m).

(e) Compare the accelerations of each object.

The slope for each object's graph is constant indicating a constant velocity,

therefore zero acceleration.

(f) Which object had the greatest displacement?

Δs3= 14.50 m - 0 = 14.50 m,Δs2= 10.00 m - 3.50 m = 6.50 m, and

Δs1= 10.00 m - 7.00 = 3.00 m.

When answering questions such as these, do not hung up on whether the answer

is 3.00 m or 3.20 m. The technique or concept isthe important thing.

3) Car 1 starts from rest after the traffic light turns green. At the same time, Car 2

passes Car 1 at 50.00 mi/h. Use the graph below to answer the following questions.

Give a reason or calculation to support your answers.

(a) How long does it take for both cars to have the same velocity?

Where graphs 1 and 2 intersect is ≈ 0.050 h.

(b) At the time in (a), how much is Car 2 ahead of Car 1?

Δs1= vΔt = ½bh = ½*0.050h*60.00 mi/h= 1.5 mi

Δs2= vΔt = bh = 0.050h*60.00 mi/h= 3.0 mi

Car 2 is ahead by 3.0 mi - 1.5 mi = 1.5 mi

Δs1= vΔt = ½bh + bh = ½*0.080h*100.00 mi/h+ 0.040*100.00 mi/h= 8.0 mi

Δs2= vΔt = bh = 0.120h*60.00 mi/h= 7.2 mi

Car 1 is ahead by 8.0 mi -7.2mi = .8 mi.

(d) How long does it take Car 1 to catch up with Car 2?

Does Car 1 catch up with Car 2 before or after 0.080 h? The answer to this

question is important to know when setting up the displacementequation for

Car 1.If Car 1 catches up withCar 2 before 0.080 h, you will not need to account

for theb*h term after 0.080 h for Car 1.

Δs1= vΔt = ½bh = ½*0.080h*100.00 mi/h= 4.0 mi

Δs2= vΔt = bh = 0.080h*60.00 mi/h= 4.8 mi

At t = 0.080 h, Car 2 is still ahead of Car 1, so the time to catch up will be greater

than 0.080 h.Therefore,

Δs1= Δs2because each car travels the same distance from the redlight.

½b1h1+ b1h1= b2h2

½*0.080h*100 mi/h+ 100 mi/h*(t - 0.080h) = 60t

t = 0.10 h

(e) When Car 1 catches up to Car 2, how far are they from the traffic light?

Use either the kinematics equation for Δs1or Δs2:

Δs1=½b1h1+ b1h1=½*0.080h*100 mi/h+ 0.020h*100 mi/h= 6.0 mi

Δs2= bh = 0.100h*60.00 mi/h= 6.0 mi

(f)For both cars, what would their Acceleration vs Time graphs look like? You need

not draw the graphs but just provide a verbal description.

For Car 1, the acceleration during the first 0.080 h, is 1250 mi/h2. At the point

1250 mi/hr2, the graph wouldbe a horizontal line from 0 h to 0.080 h. From

0.080 h to0.140 h, the graph would be a straight line on top of the x-axisfrom

0 to 0.140 h.