# Velocity, Acceleration and Displacement time graphs.

• unknown physicist
In summary: Say the acceleration is constant at 5m/s2. Then you'll write a=5. What shape does a=5...In summary, the problem involves sketching velocity vs time, acceleration vs time, and displacement vs time graphs for two different motions. The first motion is an object starting at rest and accelerating at a constant rate to a velocity of 5.0 m/s [E], then continuing at that velocity. The second motion is a car accelerating from 0 to 60 km/hr at a constant rate and then hitting a wall and slowing to a stop immediately. To solve for the shapes of the graphs, the SUVAT equations should be used and plugged in with
unknown physicist
Poster has been reminded that showing an attempt is not optional at the PF

## Homework Statement

Sketch velocity vs time, acceleration vs time and displacement vs time graphs describing the following motions.a) An object begins at rest and accelerates at a constant rate to a velocity of 5.0 m/s [E]. The object continues along at this velocity.

b) A car accelerates from 0 to 60 km/hr at a constant rate and then hits a wall slowing to a stop almost immediately.

Suvat equations

## The Attempt at a Solution

I am not sure how to solve them at all. How do I figure out the shape of the graph mathematically? I am not sure on how to start.

Hi unknown physicist and welcome to pF.

unknown physicist said:
I am not sure on how to start.
Start with the SUVAT equations that you quoted. Instead of writing the acronym, write down the actual equations and take a good look at them to see how they relate to what is being asked.

berkeman
Welcome to PF!
You are supposed to make an attempt at a solution when posting in the HW forum.
unknown physicist said:
a) An object begins at rest and accelerates at a constant rate to a velocity of 5.0 m/s [E]. The object continues along at this velocity.
Which equation would you use for this?

cnh1995 said:
Welcome to PF!
You are supposed to make an attempt at a solution when posting in the HW forum.

Which equation would you use for this?

v=u+at ?

unknown physicist said:
v=u+at ?
Right. Modify this equation by putting the value of u. What shape will the modified equation give?

cnh1995 said:
Right. Modify this equation by putting the value of u. What shape will the modified equation give?
v=u+at
u=0
v=at (mx), slope with a 0 y-intercept.

unknown physicist said:
v=at (mx), slope with a 0 y-intercept.
Yes. Until??
unknown physicist said:

cnh1995 said:
Yes. Until??

Got it, Thanks. But, what about for the acceleration and displacement time graphs? How do I know what numbers to plug in?

unknown physicist said:
Got it, Thanks. But, what about for the acceleration and displacement time graphs? How do I know what numbers to plug in?
The shape of acceleration vs time graph is mentioned in the problem itself. Read again carefully.
For displacement vs time graph, make use of the known graphs. What is the mathematical relation between displacement and velocity? Which SUVAT equation will you use?

cnh1995 said:
The shape of acceleration vs time graph is mentioned in the problem itself. Read again carefully.
For displacement vs time graph, make use of the known graphs. What is the mathematical relation between displacement and velocity? Which SUVAT equation will you use?

s=1/2(u+v)t ?

cnh1995 said:
The shape of acceleration vs time graph is mentioned in the problem itself. Read again carefully.
For displacement vs time graph, make use of the known graphs. What is the mathematical relation between displacement and velocity? Which SUVAT equation will you use?
Mentioned in the problem? I can't seem to understand that.

unknown physicist said:
s=1/2(u+v)t ?
No.
The equation should be s as a function of t. All the other terms should be known constants.

unknown physicist said:
Mentioned in the problem? I can't seem to understand that.

unknown physicist said:
An object begins at rest and accelerates at a constant rate to a velocity of 5.0 m/s

cnh1995 said:
No.
The equation should be s as a function of t. All the other terms should be known constants.

so, I should use s=ut+1/2at^2?

Also, it accelerates at a constant rate, but why does the graph need to go downward? why doesn't it go up like a slope? (for acceleration)?

unknown physicist said:
so, I should use s=ut+1/2at^2?
Yes. It is in the form y=ax+bx2. What is this shape? Put the value of u and see.
unknown physicist said:
but why does the graph need to go downward?
It doesn't. Acceleration is constant w.r.t.time. How would you show this graphically?

cnh1995 said:
Yes. It is in the form y=ax+bx2. What is this shape? Put the value of u and see.

It doesn't. Acceleration is constant w.r.t.time. How would you show this graphically?

It is a quadratic, so it should be curved.

w.r.t time?

But acceleration will be a straight line, then go straight down, to an acceleration of 0, then it will continue to be 0. That is what the answer sheet shows. I can't understand why it is straight, and why it goes down.

unknown physicist said:
But acceleration will be a straight line, then go straight down, to an acceleration of 0, then it will continue to be 0.
Yes, after v becomes 5m/s. I was talking about the acceleration period.
unknown physicist said:
It is a quadratic, so it should be curved.
Yes. Which curve is y=bx2?

cnh1995 said:
Yes, after v becomes 5m/s. I was talking about the acceleration period.

Yes. Which curve is y=bx2?
yes, I can't seem to understand the acceleration part. Why is it a straight line? then why does it go down?

cnh1995 said:
Yes, after v becomes 5m/s. I was talking about the acceleration period.

Yes. Which curve is y=bx2?

the curve will be the s=1/2ut^2
because (u)(t) = 0

unknown physicist said:
yes, I can't seem to understand the acceleration part. Why is it a straight line? then why does it go down?
It is a straight line (parallel to time axis) because it is "constant" w.r.t. time. Hence, the straight line will have zero slope.
unknown physicist said:
the curve will be the s=1/2ut^2
because (u)(t) = 0
Yes. What is the name of that curve? Look up "conic sections".

cnh1995 said:
It is a straight line (parallel to time axis) because it is "constant" w.r.t. time. Hence, the straight line will have zero slope.

Yes. What is the name of that curve? Look up "conic sections".

What do you mean by w.r.t?
and how do I understand it mathematically?

unknown physicist said:
What do you mean by w.r.t?
and how do I understand it mathematically?
w.r.t. stands for "with respect to". Say the acceleration is constant at 5m/s2. Then you'll write a=5. What shape does a=5 give?

cnh1995 said:
w.r.t. stands for "with respect to". Say the acceleration is constant at 5m/s2. Then you'll write a=5. What shape does a=5 give?
Hold on,
so for the acceleration vs time graph.
u =0
and a is a constant. ( for the first part)

How would I know what to plug for the v=u+at equations?
How would I know that the numbers i'll be plugging in is different that the ones above?

I think I can understand the displacement, and velocity graphs. Just not the acceleration ones.

cnh1995 said:
w.r.t. stands for "with respect to". Say the acceleration is constant at 5m/s2. Then you'll write a=5. What shape does a=5 give?
for acceleration, I used v =u+at
and I said u=0
a=0
and v=5

I plugged them into the equation and got 5=0? what does that mean?
and then for the second part I said u= 5 and v=5 and plugged them and got 5=5??

unknown physicist said:
How would I know what to plug for the v=u+at equations?
You should first calculate everything on paper and then plot the graphs. I am not sure I understand your question clearly.

Ah..I got what you are saying. In the actual problems, the data is not sufficient, so you can't compute anything on paper. You are just expected to draw the shapes of the graphs.

cnh1995 said:
Ah..I got what you are saying. In the actual problems, the data is not sufficient, so you can't compute anything on paper. You are just expected to draw the shapes of the graphs.

For acceleration vs time graphs, can I consider acceleration to be 0 while computing?

cnh1995 said:
Ah..I got what you are saying. In the actual problems, the data is not sufficient, so you can't compute anything on paper. You are just expected to draw the shapes of the graphs.
Yeah, visualizing them is a bit difficult, I excel in the calculation part. Interpreting them is a bit difficult.

unknown physicist said:
For acceleration vs time graphs, can I consider acceleration to be 0 while computing?
No. It is non zero when the body is accelerating and becomes zero when the body reaches a velocity of 5m/s. In fact, IMO you are not expected to compute anything on paper.

cnh1995 said:
No. It is non zero when the body is accelerating and becomes zero when the body reaches a velocity of 5m/s. In fact, IMO you are not expected to compute anything on paper.
so how do you suggest i solve them?

unknown physicist said:
so how do you suggest i solve them?
They can't be solved on paper. The data given is not sufficient. Just draw the shapes and that would be sufficient.

## 1. What is velocity?

Velocity is a measure of the rate of change of an object's position. It is a vector quantity, meaning it has both magnitude (speed) and direction. It is typically measured in meters per second (m/s) in the metric system.

## 2. How is velocity calculated?

Velocity is calculated by dividing the change in an object's displacement by the change in time. The equation for velocity is v = Δx/Δt, where v is velocity, Δx is change in displacement, and Δt is change in time.

## 3. What is acceleration?

Acceleration is a measure of the rate of change of an object's velocity. It is also a vector quantity and is typically measured in meters per second squared (m/s^2). Acceleration can be positive (speeding up), negative (slowing down), or zero (constant velocity).

## 4. How is acceleration related to velocity?

Acceleration and velocity are related through the equation a = Δv/Δt, where a is acceleration, Δv is change in velocity, and Δt is change in time. This means that acceleration is the change in velocity over time, and a change in acceleration can result in a change in velocity.

## 5. What do displacement-time graphs show?

Displacement-time graphs show the relationship between an object's displacement and time. The slope of the line on the graph represents the object's velocity, and the area under the line represents the object's displacement. A steeper slope indicates a higher velocity, and a larger area indicates a greater displacement.

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