# Acceleration and Velocity: Relationship | learn smarter (2023)

Have you ever thought about the physics and mechanics of skydiving? When you fall your acceleration changes with the rate of fall and then levels off at a certain point - why is that? The answer is terminal velocity, and that's because your acceleration varies with your speed.

This means that the increase in speed as you fall is governed by an acceleration function driven by your change in speed, and at some point you will reach your terminal speed - the speed of your fall, which you cannot exceed.

## acceleration and speed ratio

A big question all aroundaccelerationIt isHow does that relate to speed?? The answer to this is that acceleration is the derivative of velocity - meaning that acceleration is the rate of change of velocity.

Conversely, if you integrate an expression for acceleration, you get the expression for velocity.

Acceleration and velocity are both vector quantities.meaning they have magnitude and direction. This means that when considering acceleration and velocity values, sign values ​​are appended to indicate direction.

For example, if you have a negative acceleration value, the object being examined will slow down as its speed decreases.

But then what are the differences between acceleration and velocity?

## Differences in acceleration and speed

While we explored the relationship and similarities betweenspeed and acceleration, The fact is that they are different variables and so the question is how do they differ.

The acceleration follows the change in speed; as such they have different values, but can also have different signs.

For example, consider the acceleration ofa car that is slowing down, ÖAcceleration would be negative for decelerationwhen the car slows down, but thevVelocity would still be positive as you drive, just at a decreasing rate.

Now that we know how acceleration and velocity are related and how they differ, let's turn to the derivatives of acceleration and velocity.

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## Derivatives of acceleration and velocity

We know thatAcceleration is the rate of change of velocitybut we also have the relationship between velocity and displacement:Velocity is the rate of change of displacement.It means thatAcceleration is the second derivative of displacement.

This relationship also works in reverse - if you take an expression for acceleration and integrate that, you get an expression for velocity, and if you take an expression for velocity and integrate that, you get an expression for displacement. The graphic below illustrates these relationships.

Cowardly. 1. Diagram showing the relationships between displacement, velocity and acceleration.

If we examine the graph, we can see how displacement, velocity, and acceleration are related, but how can we write that mathematically?

If we have an expression for themposition of an objectgiven as $$r,$$ we can see thatthe rate is how that position changes over time,$v=\frac{dr}{dt}.$ We know that tooAcceleration is measured by how much speed changes over time.then it is given by:$a=\frac{dv}{dt}=\frac{d^2r}{dt^2}.$ These are the derived relationships we use to evaluate velocity and acceleration. As seen in the graph above, if we wanted to work in the other direction, we would simply integrate,$v=\int a\quad dt$$r=\int v\quad dt.$

We now proceed to interpret the acceleration and velocity graphs.

## Acceleration and Velocity Chart

Now that we've seen how displacement, velocity, and acceleration are related, we can visualize this by looking at velocity/acceleration-time plots.

Cowardly. 2. Diagram with acceleration and velocity-time diagrams for comparison.

Based on what we saw earlier with the relationship between velocity and acceleration, we can visualize this in the diagrams above.

Differentiating we find the slope of the velocity line. If you examine the figure above you can see this - look between $$4$$ and $$5$$ seconds on the top chart and you will see that the speed has increased from $$1 m/s$$ to \ (4 m/s) increases \ ) for one second. This gives a slope of $$3$$, which you can see at the corresponding point on the acceleration-time diagram.

Conversely, if we look between $$0$$ and $$1$$ seconds on the acceleration time graph and calculate the area bounded by the line, we will return our velocity profile. This is the technique of integration - if we integrate the acceleration (in this case the area over the line) we get the velocity. The integral response is $$-3$$ because it is at the bottom of the axes, and if we look at the velocity graph we can see that the velocity has decreased by $$3 m/s.$$ during this time has.

These relationships should help you understand how we can distinguish and integrate between displacement, velocity, and acceleration, as you saw earlier.

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## Acceleration and velocity formula

taking into accountacceleration that varies with speed,Normally we would get an expression using a velocity function.

We know that our formula for acceleration and velocity is given by $a=\frac{dv}{dt},$ and if we had a velocity function to solve we would have a question of the form $\ frac{dv } {dt}=f(v).$ This question type states that the acceleration for the question is determined by a function of velocity. How would we solve this?

The question has a term $$f(v)$$ equal to our acceleration $$(a)$$. Hence, we can hope to solve this by the method of variable separation. This means that we collect equal terms on both sides of the equals sign, integrate, and then solve for $$v$$ to get an expression for velocity $$(v)$$ in terms of time $$(t)$$ and to get any constants. Let's look at the logic behind it.

We know that acceleration is the rate of change of velocity,$\so a=\frac{dv}{dt}.$We also know that we have a function $$f(v)$$ equal to acceleration is . This leads to the expression$\frac{dv}{dt}=f(v).$FFrom here we would collect all $$v$$ terms with the $$dv$$ term and collect all $$t$$ terms with our $$dt$$ term on the other side of the equals sign.

For example, if $$f(v)=\frac{t}{v}$$ where $$t$$ is a time variable, our separation of the variables would look like this: \begin{align}a& = f( v )\\ \frac{dv}{dt}&=\frac{t}{v} \\ v dv&= tdt. \end{align} From here we would integrate and get an expression of how velocity changes over time from an acceleration expression. Let's see what this could look like with an example.

The acceleration of a particle varies with its velocity and is defined by the equation $a=3v-4.$Find the particle's velocity after 0.5 seconds if it is $$20\text{ m/s } \ ) in \(t=0.$$

Solution

Step 1. Separation of variables

\begin{align} a&=3v-4\\ \frac{dv}{dt}&=3v-4\\ \frac{1}{3v-4}dv&=1dt\end{align}

Step 2. Integrate

\begin{align} \int \frac{1}{3v-4}dv&=\int 1dt \\ \frac{\ln(|3v-4|)}{3}&=t+C\end{ alinhar}

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Step 3. Reorganize Para $$v$$.

\begin{align} \frac{\ln(|3v-4|)}{3}&=t+C \\ \ln(|3v-4|) &= 3t+C\\ 3v-4&= e^{3t+C} \\ v&=\frac{ e^{3t+C} +4}{3}\end{align}

Step 4. Apply initial conditions

We know that in $$t=0\text{ s}$$ the velocity is $$20\text{ m/s}$$, i.e. $$v(0)=20.$$ By inserting it into our equation we receive ,

\begin{align}v&=\frac{ e^{3t+C} +4}{3}\\ 20&=\frac{ e^{3(0)+C} +4}{3} \\ 60&=e^C+4\\ \ln(56)&=C\end{align}

For this reason,

\begin{align}v=\frac{ e^{3t+\ln(56)} +4}{3}\end{align}

Step 5. Find the speed in $$t=0.5s$$

\begin{align}v&=\frac{ e^{3t+\ln(56)} +4}{3} \\ &=\frac{ e^{3(0.5)+\ln(56)} + 4}{3}\\ &= 84,99\text{ m/s} \end{align}

### terminal speed or limit

we can count toothe final speedan object with the same expression format as shown above.

If we take the terminal velocity as the maximum velocity that the object can reach, it means that the rate of change of velocity is 0. The object has reached its maximum speed, so the speed cannot increase further.

An example of this could be an object in free fall. At some point, the object's weight will limit the speed at which it falls, and that speed cannot be exceeded. We will also express the acceleration $$a$$ as a function of the velocity $$v$$, which leads to the general form $\frac{dv}{dt}=f(v)$$0 = f( v) .$If we then switch the function to $$v$$, we havethe limit speed. Let's look at that as an example.

An object is in free fall and its acceleration is given as:$a=\frac{237-6v^2}{24}.$Find the terminal velocity of the object.

Solution

Step 1. Write the equation in terms of rate of change

\begin{align}a&=\frac{237-6v^2}{24}\\ \frac{dv}{dt}&=\frac{237-6v^2}{24}.\end{align }We know that at terminal velocity the rate of change of velocity is 0, so$0=\frac{237-6v^2}{24}$

Step 2. Solve for $$v$$

We know that the left side of our equation must be 0. Therefore, the numerator of the fraction must be 0, so dividing it by the denominator (24) will return 0. That means $$6v^2$$ must equal 237.

\begin{align}6v^2&=237\\ v&=\sqrt{\frac{237}{6}} \\&=6.28\text{ m/s}\end{align}

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## Equations of Acceleration and Velocity

We also have another method that we can use when tracking acceleration at variable rates. We can apply thatSUVAT equationsto a problem with constant acceleration and variable speed. Regarding the notation, we have \begin{align} s&=\mbox{displacement}\\u&=\mbox{initial velocity}\\v&=\mbox{final velocity}\\a&=\mbox{acceleration} \ \ t&=\mbox{Time}\end{align} We then have the five equations for SUVAT calculations: \begin{align} &(1)\quad s= ut+\frac{at^2}{ 2 }\\&(2)\quad s= vt-\frac{at^2}{2}\\&(3)\quad s=\frac{v+u}{2}t\\&(4 ) \quad v^2=u^2+2as\\&(5)\quad v=u+at\end{align}

We can see how equations (1), (2), (4) and (5) have an acceleration term and as such the relevant rearranged equations for acceleration are:\begin{align} &( 1)\ Quad a= \frac{2s-2ut}{t^2}\\&(2)\quad a= - \left[\frac{2s-2vt}{t^2}\right]\\& (4) \quad a=\frac{v^2-u^2}{2s}\\&(5)\quad a=\frac{v-u}{t}.\end{align}

What acceleration does an object experience if it starts at $$5\text{ m/s}$$ after it has traveled $$10$$ seconds after a distance of $$37\text{ m}$$?

Solution

We know the initial velocity, time and distance, and we want to know the acceleration. This means we can use equation (1) above, that is, rearrange $s= ut+\frac{at^2}{2}$our unknown acceleration and solve:

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\begin{align}a&= \frac{2s-2ut}{t^2}\\&=\frac{(2\cdot 37)-(2\cdot 5\cdot 10)}{10^2} \\&=-0,26\text{m/s}^2\end{align}

This means that after $$10$$ seconds after the start, the object will slow down with $$0.26\text{ m/s}^2$$.

## Acceleration and Velocity - Main Conclusions

• You can switch between displacement, velocity, and acceleration by differentiating or backward integrating along that order.
• The acceleration and velocity curves are related so that the slope of the velocity curve represents the acceleration value and the area under the acceleration curve for a time interval represents the change in velocity.
• If you have an equation that gives acceleration as a function of velocity, you can use the technique of separating variables and integrating to get an expression for velocity.
• The terminal velocity of an object can be calculated by setting the rate of change of velocity to $$0$$ and solving for $$v.$$. This has the general form $$0=f(v).$$

## FAQs

### What is the relationship between velocity and acceleration? ›

Acceleration is the rate of change of velocity. (when velocity changes -> acceleration exists) If an object is changing its velocity, i.e. changing its speed or changing its direction, then it is said to be accelerating. Acceleration = Velocity / Time (Acceleration)

How are velocity and acceleration related for kids? ›

Acceleration is the measurement of change in an object's velocity. When you press down on the gas pedal in a car, the car surges forward going faster and faster. This change in velocity is acceleration. The standard unit of measurement for acceleration is meters per second squared or m/s2.

What is the difference between velocity and acceleration? ›

Difference between Velocity and Acceleration

Velocity is the rate of change of displacement. Acceleration is the rate of change of velocity. Velocity is a vector quantity because it consists of both magnitude and direction. Acceleration is also a vector quantity as it is just the rate of change of velocity.

What is the relationship between velocity and acceleration quizlet? ›

Velocity is how fast an object moves and acceleration is the rate of change in velocity.

What happens when velocity and acceleration are the same? ›

If the velocity and acceleration are in the same direction (both have the same sign - both positive or both negative) the object is speeding up. If the velocity and acceleration are in opposite directions (they have opposite signs), the object is slowing down.

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