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Eva8 [605]
3 years ago
6

How can you tell when an object changes direction on velocity vs time graph

Physics
2 answers:
jek_recluse [69]3 years ago
8 0

On velocity vs time graphs, if an object's velocity goes from positive to negative, or negative to positive, then it has changed directions.

A positive velocity means the object is traveling in the positive direction and a negative velocity would mean it has turned around and begun traveling in the opposite, or negative direction.

Grace [21]3 years ago
3 0

In my opinion, that is one humungously fantastic question, that is not asked nearly often enough.

The answer is:. You can't ! None of the so-called "velocity/time" graphs shows any information about the direction of the motion, and there's no way to tell if it changes, from the graph.

There's no difference between the "velocity" graphs of an object in straight-line motion with constant speed, and another object in uniform circular motion. Both graphs are horizontal lines.

A "velocity" graph can't show direction OR displacement, and the way I see it, such a graph is not entitled to use the 5-dollar word 'velocity' at all. It is simply a graph of SPEED vs time, and no more.

So there !

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5 0
3 years ago
Read 2 more answers
A projector is placed on the ground 22 ft. away from a projector screen. A 5.2 ft. tall person is walking toward the screen at a
Stella [2.4K]

Answer:

y = 67.6 feet,   y = 114.4/ (22 - 3t)

Explanation:

For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram

Large triangle Projector up to the screen

         tan θ = y / L

For the small triangle. Projector up to the person

         tan θ = y₀ / (L-d)

The angle is the same, so we equate the two equations

         y₀ / (L -d) = y / L

         y = y₀  L / (L-d)

The distance from the screen (d), we look for it with kinematics

         v = d / t

        d = v t

we replace

         y = y₀ L / (L - v t)

         y = 5.2 22 / (22 - 3 t)

         y = 114.4 (22 - 3t)⁻¹

This is the equation of the shadow height change as a function of time

For the suggested distance the shadow has a height of

           y = 114.4 / (22-13)

           y = 67.6 feet

7 0
3 years ago
What is the calculation for elastic potential?
grin007 [14]
Elastic potential energy is equal to the force times the distance of movement. Elastic potential energy = force x distance of displacement. Because the force is = spring constant x displacement, then the Elastic potential energy = spring constant x displacement squared.
3 0
3 years ago
What is the minimum amount of data points an experiment should gather?
nydimaria [60]

The answer is "Three".

At the point when individuals take studies for factual purposes or when the Statistics Agency comes around and gets everyone's data, data from one individual is one information point for them. Toward the finish of its exploration or overview, the organization will have accumulated numerous bits of data from numerous individuals. One piece of information equals with one data point.

7 0
3 years ago
At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzl
Debora [2.8K]

Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

s=ut+0.5at^2 \\

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\

Solving both equations

600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\

Passing of B occurs at 4108.31 height.

6 0
3 years ago
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