**Instantaneous velocity** tells us about the motion of a particle at a specific instant of time anywhere along its path.

**Instantaneous velocity** **is taken as the limit of average velocity as the time tends towards zero.** **To Calculate V _{inst} we can use the displacement-time graph/ Instantaneous Velocity Formula**.

**i.e.,**

**the derivative of displacement (s) with respect to time(t) taken**.

[latexpage]

\[\vec{V_{inst}}=\lim_{\Delta t\rightarrow 0}\frac{\Delta\vec {s} }{\Delta t}=\frac{d \vec{s}}{dt}\]

**To know how to calculate instantaneous velocity of an object, we have steps to follow**. **Let us see it with an example. **

**Consider an equation for velocity in terms of position/displacement.**

To calculate **instantaneous velocity**, we must consider an **equation** that tells us its **position** **‘s’ **at a certain **time ‘t’**. It means the equation must contain the variable ‘**s**‘ on one side and ‘**t**‘ on the other side,

s = -2tat t = 2 second.^{2}+ 10t +5^{ }

In this equation, the variables are:

**Displacement = s, **measured in meters.

**Time = t**, measured in seconds.

**Consider the derivative of the given equation.**

To find the derivative of a given displacement equation, **differentiate the function with respect to time,**

ds/dt= -(2) 2t^{(2-1)}+ (1)10t^{1 – 1}+ (0)5t^{0}

ds/dt= -4t^{1}+ 10t^{0}

ds/dt = -4t + 10

**Substitute the given value of “t” in the derivative equation to find instantaneous velocity.**

Find the **instantaneous velocity** at t = 2, substitute **“2”** for t in the derivative ds/dt = -4t + 10. Then, we can solve the equation,

ds/dt = -4t + 10

ds/dt = -4(2) + 10

ds/dt = -8 + 10

ds/dt = -2 meters/second

*Here, “meters/second” is the SI Unit of instantaneous velocity.*

**How to calculate instantaneous velocity from a graph**

**Instantaneous velocity at any specific point of time is given by the slope of tangent drawn to the position-time graph at that point.**

- Plot a graph of
**distance vs. time.** - Mark a point at which you have to find instantaneous velocity, say
**A.** - Determine the point on the graph corresponding to time
**t**_{1}_{ }and**t**_{2}. - Calculate the
**v**and draw a tangent at point_{avg}**A**. - In the graph,
**v**at point_{inst}**A**is found by tangent, drawn at that point

- Longer the tangent, the more accurate will be the values.
- In the image shown, blue line is the
**position vs. time graph**, and the red line is an approximated slope for the line at t = 2.5 seconds.

- If we keep choosing points that are closer and closer to one another, the line will begin approaching the slope of the line tangent to a single point.
- If we take the limit of the function at that point, we will get the value of the slope of tangent at that point.
- The distance is approximately 140 m, and the time interval is 4.3s. Therefore, the approximated slope is 32.55 m/s.

**How to calculate instantaneous velocity from a position-time graph.**

**To calculate the instantaneous velocity from a position-time graph.**

**Plot the displacement function with respect to time.**

- Use the x-axis and y-axis to represent
**time and displacement**. - Then plot the values of time and displacement on the graph.

**Choose any two points on the s-t graph.**

- The displacement line contains the points (3,6) and (5,8).
- In this example, if we want to find slope at (3,6), we can set
**A = (3,6)**and**B****=****(5,8)**

** **

**Find the line slope connecting the two points, i.e., between A and B.**

Find the average velocity between those two-time intervals, i.e.,

[latexpage]

\[slope=\textbf{K}=\frac{Y_{B}- Y_{A}}{X_{B}-X_{A}}\]

where K is the slope between the two points.

Here, the slope between A and B is:

[latexpage]

\[slope=\textbf{K}=\frac{(8-6)}{(5-3)}=2\]

**Repeat to find slope several times, moving B nearer to A.**

- Keep choosing points closer to one another; then, it will begin to approach the slope of the tangent line.
- If we consider the limit of the function at that point, we will get the value of the slope at that point.
- Here we can use the points (4,7.7), (3.5, 6.90), and (3.25, 6.49) for B and the original point of (3,6) for A.

- At B = (4,7.7)

[latexpage]

\[slope=\textbf{K}=\frac{(7.7-6)}{(4-3)}=1.7\]

- At B = (3.5, 6.90)

[latexpage]

\[slope=\textbf{K}=\frac{(6.90-6)}{(3.5-3)}=1.8\]

- At B = (3.25, 6.49)

[latexpage]

\[slope=\textbf{K}=\frac{(6.49-6)}{(3.25-3)}=1.96\]

**Calculate the slope for an infinitely small interval on the tangent line.**

In the example, as we move B closer to A, we get values of 1.7, 1.8, and 1.96 for **K**. Since these numbers are approximately equal to 2, we can say that **2** is A’s slope.

Here, **instantaneous velocity is 2m/s.**

**Instantaneous velocity formula**

In mathematical terms, we can write the **instantaneous velocity formula** as,

[latexpage]

\[Instantaneous{\enskip} velocity=\frac{Change{\enskip} in {\enskip} position}{Time {\enskip} interval}\]

[latexpage]

\[\vec{V_{inst}}=\lim_{\Delta t\rightarrow 0}\frac{\Delta\vec {s} }{\Delta t}=\frac{d \vec{s}}{dt}\]

Here, **ds/dt is the derivative of displacement (s) with respect to time (t).**

The above **derivative holds a finite value** when both the denominator and the numerator tend to zero.

**Instantaneous velocity formula calculus**

**By using calculus, it is always possible to calculate the velocity of an object at any moment along its path. It is called instantaneous velocity **and is given by the equation v = ds/dt.

**Instantaneous velocity = limit as change in time approaches zero** (change in position/change in time)** = derivative of displacement with respect to time**

[latexpage]

\[\vec{V_{inst}}=\lim_{\Delta t\rightarrow 0}\frac{\Delta\vec {s} }{\Delta t}=\frac{d \vec{s}}{dt}\]

\[V_{inst}=\lim_{\Delta t\rightarrow 0}\frac{ds}{dt}\]

\[\vec{V}= instantaneous{\enskip} velocity\]

\[\Delta {\vec{S}} = vector{\enskip} change{\enskip} in {\enskip}position(m)\]

\[\Delta {t} =change{\enskip} in {\enskip}time(s)\]

\[\frac{ds} {dt}=derivative{\enskip} of{\enskip} position{\enskip} vector{\enskip} with {\enskip}respect {\enskip}to {\enskip}time (m/s)\]\[s = displacement\]\[t = time\]

**Average velocity and instantaneous velocity formula**

Formula | Symbol | Definition | |

Average velocity | s= Final _{f} displacement s_{i}_{ = }Initial displacementt_{f }_{=} Final timet = Initial time_{i} | Average velocity is total distance divided by the total time taken. | |

Instantaneous velocity | Velocity at any instant of time. |

**Instantaneous angular velocity formula**

The **instantaneous angular velocity** is the rate at which a particle moves in a circular path at a particular moment of time.

The **instantaneous angular velocity** of a rotating object is given by

[latexpage]

\[\omega_{av}=\lim_{\Delta t\rightarrow 0}\frac{\Delta\theta}{\Delta t}=\frac{d \theta}{dt}\]

` `

** dθ/dt**** ** = **derivative of angular position θ with respect to time**, found by taking the limit Δ t → 0 in the

**average angular velocity**.

[latexpage]

\[\omega_{av}=\frac{\theta_{2}- \theta_{1}}{t_{2}-t_{1}}= \frac{\Delta{\theta}}{\Delta t}\]

The **direction of the angular velocity in a circular path is along the axis of rotation** and points away from you for a body rotating **clockwise** and toward you for a body rotating **anticlockwise**. In mathematics, this is generally described by the **right-hand rule.**

**Instantaneous velocity and speed formula**

**The formula of instantaneous velocity**

[latexpage]

\[\vec{ V_{inst}}=\lim_{\Delta t\rightarrow 0}\frac{\Delta\vec {s} }{\Delta t}=\frac{d \vec{s}}{dt}\]

**The formula for instantaneous speed**

[latexpage]

\[speed_{inst}=\frac{ds}{dt}\]

*Difference between Instantaneous Speed and Instantaneous Velocity.*

*Difference between Instantaneous Speed and Instantaneous Velocity.*

Instantaneous velocity | Instantaneous speed |

It is the velocity of a particle in motion at a particular moment of t. | It is the measure of speed of a particle at a specific moment of t. |

Instantaneous velocity measures how fast and in which direction an object is moving. | Instantaneous speed measures how quickly a particle is moving in motion. |

Vector quantity | Scalar quantity |

**Instantaneous velocity definition and formula**

**Instantaneous velocity definition**

**Instantaneous velocity **is described as the velocity of an object in motion. We can find it by using average velocity, but we must narrow the time to approach zero.

In total, we can say that **instantaneous velocity is the velocity of a particle in motion at a particular instant of time.**

**Instantaneous velocity formula**

For any equation of motion *s*(*t*), for **instantaneous velocity** as t approaches zero, we can write the **formula** as,

[latexpage]

\[\vec{ V_{inst}}=\lim_{\Delta t\rightarrow 0}\frac{\Delta\vec {s} }{\Delta t}=\frac{d \vec{s}}{dt}\]

**Instantaneous velocity** **limit formula**

**The instantaneous velocity of any object is the limit of the average velocity as the time approaches zero**.

[latexpage]

\[Instantaneous {\enskip}velocity = v = \frac{\Delta s(t)}{\Delta t}\]

\[Instantaneous{\enskip} velocity=\vec{ V_{inst}}=\lim_{\Delta t\rightarrow 0}\frac{\Delta\vec {s} }{\Delta t}\]

\[\vec{ V_{inst}}=\lim_{\Delta t\rightarrow 0}\frac{\Delta\vec {s} }{\Delta t}=\frac{s(t_{2})- s(t_{1})}{t_{2}-t_{1}}\]

\[\vec{ V_{inst}}=\lim_{\Delta t\rightarrow 0}\frac{\Delta\vec {s} }{\Delta t}=\frac{s(t + \Delta t)- s(t)}{(t+{\Delta t})-t}\]

\[\vec{ V_{inst}}=\lim_{\Delta t\rightarrow 0}\frac{\Delta\vec {s} }{\Delta t}=\frac{s(t + \Delta t)- s(t)}{\Delta t}\]

Insert the values of t_{1}= t and t_{2} = t + Δt into the equation for the average velocity and take the limit as Δt→0, we find **the instantaneous velocity limit formula**

**How do you find instantaneous velocity on a graph**

**Instantaneous velocity equals the slope of tangent line of the position-time graph.**

**Instantaneous Velocity interpretation from s-t graph**

**Instantaneous velocity equals the slope of tangent line of the position-time graph.****Instantaneous Velocity interpretation from s-t graph**

[latexpage]

\[\vec{ V_{inst}}=\lim_{\Delta t\rightarrow 0}\frac{\Delta\vec {s} }{\Delta t}=\frac{d \vec{s}}{dt}\]

- The slope of the purple line (tangent) in the displacement v/s time graph gives instantaneous velocity.
- If the purple line makes an angle with the positive x-axis.

` V`

= slope of purple line = _{inst} `tanθ`

` `

**How do you find instantaneous velocity from average velocity**

To find the **instantaneous velocity at a point, we have to first find the average velocity at that point.**

You can find the instantaneous velocity at t=a by **calculating the average velocity of the position vs. time graph by taking the smaller and larger increments of a point at which you want to determine V**_{inst}_{.}

**Instantaneous velocity example**

While riding his bicycle, a cyclist **changes his velocity depending on the distance and time he travels.**

If we want to find the velocity at one particular point, we must use instantaneous velocity.

Let us see **an example,**

## **a). Find out the Instantaneous Velocity of a particle traveling along a straight path for t=2 seconds, with a position function “s” defined as 4t² + 2t + 3?**

**Solution:**

Given **s = 4t² + 2t + 3**

Differentiate the given function with respect to time, we calculate the Instantaneous Velocity as follows:

Substitute value of t = 2, we get the instantaneous velocity as,

[latexpage]

\[v_{inst} = \frac{ds}{dt}\]

Substituting function s,

[latexpage]

\[v_{inst} = \frac{d(4t^2 +2t +3)}{dt}} \]

\[v_{inst} =8t+2\]

\[v_{inst} = (8 * 2)+2\]

\[v_{inst} =18 ms ^{-1}\]

Thus, the instantaneous velocity for the above function is 18 m/s.

**Instantaneous velocity problem**

**Some Instantaneous velocity problems,**

**Problem 1:**

**The motion of the truck is given by the function s = 3t**^{2} + 10t + 5. Calculate its Instantaneous Velocity at time t = 4s.

^{2}+ 10t + 5. Calculate its Instantaneous Velocity at time t = 4s.

**Solution:**

Given function is s = 3t^{2 }+ 10t + 5.

Differentiate the above function with respect to time, we get

[latexpage]

\[{v_{inst} = \frac{ds}{dt}=\frac{d(3t^2 +10t +5)}{dt}}\]

Substituting function s,

[latexpage]

[v_{inst} = v(t)=6t+10]

Substitute value of t = 4s, we get the instantaneous velocity as,

[latexpage]

\[v(4)= 6(4)+10\]

\[v(4) =34 ms ^{-1}\]

For the given function, Instantaneous Velocity is 34m/s

**Problem 2:**

**A bullet fired travels along a straight path, and its equation of motion is S(t) = 3t + 5t**^{2}. So, for example, if it travels for 12 seconds before impact, find the instantaneous velocity at t = 7s.

^{2}. So, for example, if it travels for 12 seconds before impact, find the instantaneous velocity at t = 7s.

**Solution:** We know the equation of motion:

[latexpage]

\[s(t) = 3t + 5t^2\]

\[v_{inst} = \frac{ds}{dt} = \frac{d(3t +5t^2)}{dt}=3+10t}\]

\[v_{inst} {\enskip} at (t = 7) = 3 + (10 * 7)\]

\[v_{inst} = 73 m/s.\]

**Problem 3:**

**An object is released from a certain height to fall freely under the influence of gravitation. The equation of motion for displacement is s(t) = 5.1 t**^{2}. What will be the instantaneous velocity of an object at t=6s after release?

^{2}. What will be the instantaneous velocity of an object at t=6s after release?

**Solution:**

The equation of motion is

s(t) = 5.1 t^{2}

Instantaneous velocity at t = 6s

[latexpage]

\[v_{inst} = [\frac{ds}{dt}]_{t=6} = [\frac{d(3t +5t^2)}{dt}]_{t=6}=3+10t}\]

\[v_{inst} = [5.1 * 2 * t]_{t=6}\]

\[v_{inst} = [5.1 * 2 * 6]\]

\[v_{inst} = 61.2 ms ^{-1}\]

**Problem 4:**

**Find the velocity at t = 2, given the** **displacement equation is s = 3t**^{3} – 3t^{2} + 2t + 7.

^{3}– 3t

^{2}+ 2t + 7.

**Solution:**

It is just like previous problems, except they have given a cubic equation instead of a quadratic equation to solve it in the same way.

The equation of motion is

s(t) = 3t^{3} – 3t^{2} + 2t + 7.

[latexpage]

\[v_{inst} = \frac{ds}{dt} = \frac{d(3t^3+3t^2 +2t+7)}{dt}=(3*3t^2) – (2 * 3t) + 2}\] \[v_{inst} = [9t^2-6t+2]\]

Instantaneous velocity at t = 7s

[latexpage]

\[v_{inst} = 9(7)^{2} – 6(7) +2\]

\[v_{inst} = 441 – 42 +2\]

\[v_{inst} = 401{\enskip} meters/second\]

** Problem 5:**

**The position of a person moving along a straight line is given by s(t)= 7t**^{2}+ 3t + 19, where t is time(seconds). Find the equation for instantaneous velocity v(t) of the particle at time t.

^{2}+ 3t + 19, where t is time(seconds). Find the equation for instantaneous velocity v(t) of the particle at time t.

**Solution:**

Given: s(t)= 7t^{2}+ 3t + 19

[latexpage]

\[v_{inst} = \frac{ds}{dt} = \frac{d(7t^2 +3t+19)}{dt}\]

\[v_{inst} = 14t+ 3\]

**v _{inst} = v(t) = (14t + 3) m/s is equation for instantaneous velocity.**

Suppose if we assume t = 3s, then

[latexpage]

\[v_{inst}=v(t) = [14(3) + 3)] = 45 m/s\]

**Problem 6:**

**The motion of an auto is described by the equation of motion s = gt**^{2} + b, where b=20 m and g = 12 m. Therefore, find the instantaneous velocity at t=4s.

^{2}+ b, where b=20 m and g = 12 m. Therefore, find the instantaneous velocity at t=4s.

**Solution:**

s(t) = gt^{2} + b

v (t) = 2gt + 0

v (t) = 2gt

Here, g = 12 and t = 4s,

v (4) = [2 x 12 x 4] = 96 m/s.

**v (t) **= **96 m/s.**

**Problem 7:**

**A table dropped off a 1145 ft building, has a height (in feet) above the ground is given by s(t) = 1145 -12 t**^{2}. Then, calculate the instantaneous velocity of the table at 3s?

^{2}. Then, calculate the instantaneous velocity of the table at 3s?

**Solution:**

[latexpage]

\[ V_{inst}=\lim_{\Delta t\rightarrow 0}\frac{\Delta {s} }{\Delta t}=\frac{s(t_{2})- s(t_{1})}{t_{2}-t_{1}}\]

\[ V_{inst}=\lim_{\Delta t\rightarrow 0}\frac{\Delta {s} }{\Delta t}=\frac{s(t + \Delta t)- s(t)}{(t+{\Delta t})-t}\]

\[ V_{inst}=\lim_{a \rightarrow 0}\frac{[1145 – 12((t + \Delta t)^{2}]-[1145-12(t)^{2}]} {\Delta t}\]

\[consider{\enskip} \Delta {t} = a {\enskip} and {\enskip} t =3s\]

\[ V_{inst}=\lim_{a \rightarrow 0}\frac{[1145 – 12(3 + a)^{2}]-[1145-12(3)^{2}]} {a}\]

\[ V_{inst}=\lim_{a \rightarrow 0}\frac{[1145 – 12(3^{2} + a ^{2} + 6a]-[1145-12(9)]} {a}\]

\[ V_{inst}=\lim_{a \rightarrow 0}\frac{[1145 – 108 – 12a ^{2} – 72a]-1145 + 108]} {a}\]

\[ V_{inst}=\lim_{a \rightarrow 0}\frac{ – 12a ^{2} – 72a} {a}\]

\[ V_{inst}=\lim_{a \rightarrow 0}\frac{ – 12a – 72} {1}\]

\[ V_{inst}= -72 m/s\]

**Instantaneous velocity at t = 3s is -72m/s.**

**Problem 8:**

**A particles position function is given by s = (3t**^{2})_{i} – (4t)_{k} + 2. what is its instantaneous velocity at t=2? What is its instantaneous acceleration as a function of time?

^{2})

_{i}– (4t)

_{k}+ 2. what is its instantaneous velocity at t=2? What is its instantaneous acceleration as a function of time?

**Solution:**

s(t) = ** **(3t^{2})_{i} – (4t)_{k} +2

v (t) = (6t)_{i} – 4_{k}…………..(Eq.1)

v (2) = (6 * 2)_{i} – 4_{k}

v (2) = 12_{i} – 4_{k} m/s

To calculate instantaneous acceleration as a function of time

a (t) = v^{1}(t)

differentiate Eq.1 w.r.to t, we get

a (t) = 6_{i} m/s

**Problem 9:**

**The position of an insect is given by s = 44 + 20t – 3t**^{3}, where t is in seconds and s is in meters.

^{3}, where t is in seconds and s is in meters

## **a. Find the average velocity of object between t = 0 and t = 4 ****s**.

## **b. At what time between 0 and 4 is the instantaneous velocity zero**.

**solution:**

**To calculate average velocity**

[latexpage]

\[\vec{ V_{avg}}=\frac{d\vec {s}}{dt}=\frac{s_{f}- s_{i}}{t_{f}-t_{i}}=\frac{s(4)- s(0)}{4-0}\]

\[\vec{v_{avg}}= \frac{[44 + 20(4) – 3 (4)^{3} ] – 44]}{4}\]

\[\vec{v_{avg}} = -28 m/s\]

**To find the time at which instantaneous velocity is zero.**

[latexpage]

\[\vec{ V_{inst}}=\frac{d\vec {s}}{dt}= 20-9 t^{2}\]

[latexpage]

\[20-9 t^{2}=0\]

\[ t = \sqrt {20}{9}\]

\[t = 1.49 s\]

**Problem 10:**

**A particle is in motion with displacement function s = **** t**^{2} + 3.

^{2}+ 3

**Find the position at t = 2.**

**Find average velocity from t = 2 to t = 3.**

**Find its instantaneous velocity at t = 2**.

**Solution:**

**To find position at t = 2**

s(t) = t^{2} + 3

s (2) = (2)^{2} + 3

**s (2) = 7**

**To find the** **average velocity.**

[latexpage]

\[\vec{ V_{avg}}=\frac{\Delta\vec {s}}{\Delta t}\]

\[\vec{ V_{avg}}=\frac{s_{f}- s_{i}}{t_{f}-t_{i}}=\frac{s(12)- s(7)}{3-2}= 5 m/s\]

**To find instantaneous velocity**

[latexpage]

\[\vec{ V_{inst}}=\frac{d\vec {s}}{dt}\]

\[\vec{ V_{inst}}= 2t\]

** At t = 2s**

[latexpage]

\[\vec{ V_{inst}} = 2(2) = 4 m/s \]

**Instantaneous velocity vs. average velocity**

Instantaneous velocity | Average velocity |

The instantaneous velocity is the average velocity between two points. | Average velocity is the ratio of change of distance with respect to time over a period. |

Instantaneous velocity tells about the motion between two points on the path taken. | Average velocity does not give information about motion between the points. The path may be straight/curved, and the motion may be steady/variable. |

Instantaneous velocity is equal to slope of the tangent of displacement(s) vs. the time graph. | It equals to the slope of the secant line of the s-t graph. |

vector | vector |

**How to find** **instantaneous velocity without calculus**

**W**e can **find instantaneous velocity** by approximation on the **displacement vs. time graph** without calculus at a particular point. We need to draw a tangent at a point along the curved line and estimate the slope where you need to find instantaneous velocity.

**How do I calculate instantaneous velocity and instantaneous acceleration**

| Instantaneous velocity | Instantaneous acceleration |

From formula | To calculate Instantaneous velocity, take the limit of change of distance with respect to time taken as time approaches zero. i.e., by taking the first derivate of the displacement function.
| To calculate Instantaneous acceleration, take the limit of change of velocity with respect to time as the change in time approaches zero. i.e., by taking the second derivate of the displacement function. |

From graph | Equal to slope of the tangent of the s-t graph. | Equal to slope of the tangent of the v-t graph. |

## **Problem 11:**

**A bullet fired in space travels along a straight path, and its equation of motion is s(t) = 2t +**^{ }4t^{2}. If it travels for 12 seconds before impact, find the instantaneous velocity and instantaneous acceleration at the t = 3s.

^{ }4t

^{2}. If it travels for 12 seconds before impact, find the instantaneous velocity and instantaneous acceleration at the t = 3s.

**Solution:** We know the equation of motion: s(t) = 2t + 4t^{2}

[latexpage]

\[v_{inst}= \frac{ds}{dt} = \frac{d(2t+4t^{2})}{dt} = 2+ 8t\]

\[v_{inst} {\enskip}at{\enskip} v(t=7) = 2 + (8 * 3)\]

\[v_{inst}=26m/s\]

[latexpage]

\[a(t)= \frac{dv}{dt} = \frac{d(2+8t)}{dt} = 8\]

\[a(t)=8 m/s\]

**How to find instantaneous speed and velocity**

**Instantaneous speed is given as the magnitude of the instantaneous velocity.**

If displacement as a function of time is known, we can find out the instantaneous speed at any time.

Let’s understand this by an example.

**Problem 12:**

**Equation of motion is s(t) = 3t ^{3} **

[latexpage]

\[Instantaneous{\enskip} speed = \frac{ds}{dt} \]

\[s_{inst} = \frac{d(3t^{3})}{dt} = 9t^{2}\]

Consider t = 2s

[latexpage]

\[s_{inst} = 9(2)^{2}= 36m/s \]

**Why is it possible to calculate instantaneous velocity using kinematic formulas only when acceleration is constant**

**Kinematics equations can be used only when the acceleration of the object is constant.**

In the case of **variable accelerations**, Kinematics equations will be different depending on the function form the acceleration takes; at that time; we should use the **integrated approach** to calculate **instantaneous velocity**. Which will be a bit complex.

**Why do we take small time intervals while calculating instantaneous velocity. How does it give velocity at that instant if we are calculating it over a certain time interval**

The **instantaneous velocity** is given by,

\[\vec{ V_{inst}}=\lim_{\Delta t\rightarrow 0}\frac{\Delta\vec {s} }{\Delta t}=\frac{d \vec{s}}{dt}\]

The smaller the value of “**t**” , the more closely will be the **slope of the tangent line, i.e., instantaneous velocity.**

When you want to **calculate the velocity** at a specific time, you need to first calculate the **average velocities **by taking small intervals of time. If those average velocities give the same value, then it will be the required **instantaneous velocity.**

**Is velocity and instantaneous velocity different**

**Instantaneous velocity is different from velocity.**

**Velocity **is generally known as the rate of change of position with time. In contrast, in **instantaneous velocity**, the time interval is narrowed to approach zero to give velocity at a particular instant of time.

**For example,**

A particle **moving in a circle has zero displacements**, and it is required to know the velocity of a particle. In this case, we can calculate instantaneous velocity because it has a **tangential velocity** at any given point of time.

**What is instantaneous velocity with real-life examples**

**Instantaneous velocity real-life examples**

If we consider an example of a squash ball, the ball comes back to its initial point; at that time, the total displacement and average velocity will be zero. In such cases, the motion is calculated by **instantaneous velocity**.

- The speedometer of a vehicle gives information about
**the instantaneous velocity/speed of**a vehicle. It shows velocity at a particular instant of time.

- In a race, photographers take snapshots of runners, their average velocity doesn’t change, but their instantaneous velocity, as captured in the “snapshots,” changes. So it will be an
**example of instantaneous velocity.**

- If you are near a shop and a vehicle crossed in front of you at “
**t**“second, and you start to think about its velocity at a particular*time*, here you would be referring to the**instantaneous velocity of the vehicle.**

**Frequently asked questions | FAQs**

**Is instantaneous velocity a vector**

Instantaneous velocity is a vector quantity.

**Instantaneous velocity is a vector because it has both magnitude and direction. It shows both speed (refers to magnitude) and direction ****of a partic****le. It has a dimension of LT ^{-1}.We can determine it by taking the slope of the distance-time graph**.

**How do you find instantaneous velocity with only a position vs. time graph and without an equation given**

We can determine instantaneous velocity by taking the slope of the position-time graph.

**Plot a graph of displacement over time.****Choose point A and another point B that is near to A on the line.****Find the slope between A and B, calculate several times, moving A nearer to B.****Calculate the slope for an infinitely small interval on the line.****The slope obtained is instantaneous velocity.**

**Is it possible to instantaneously change velocity**

It is not possible to bring an instantaneous change in velocity since it would require infinite acceleration.

**In general, acceleration is the result of F = ma**

[latexpage]

\[a= \frac{F}{m} = (Force{\enskip} over{\enskip} a {\enskip}mass)\]

**and velocity is the outcome of the acceleration (from integration).If a change in velocity is a step function and as the time approaches zero, it would require infinite acceleration and force to change the velocity of mass instantaneously.**

**How can I calculate displacement when acceleration is a function of instantaneous velocity Initial velocity is given**

We can calculate displacement in two ways, when Initial velocity is given

**From derivation**

**Here acceleration is a function of instantaneous velocity, **

[latexpage]

\[a =\frac{dv}{dt}\]

**Initial velocity**

[latexpage]

\[v = \frac{ds}{dt}\]

\[a = \frac{d(ds)}{dt^{2}}\]

\[d(ds) = a dt^{2}\]

**By integrating,**

[latexpage]

\[ds =\int {a dt^{2}\]

**Using this form, you can get displacement ds.**

**From the formula**

**By using the below kinematic equation, we can find displacement,**

[latexpage]

\[S = ut + \frac{1}{2} at^{2}\]

**What is average and** **instantaneous velocity**

The average velocity and instantaneous velocity are expressed as follows,

Average velocity | Instantaneous velocity |

The average velocity for a particular time interval is total displacement divided by total time. | Both time interval and displacement approach zero at some point. But the limit of the derivative of displacement to total time interval is non-zero, called instantaneous velocity. |

Average velocity is the velocity of the whole path in motion | while instantaneous velocity is the velocity of a particle at a specific time |

v =` s/t` | v = `ds/dt` |

**Is instantaneous acceleration perpendicular to instantaneous velocity**

Instantaneous acceleration of the body is always perpendicular to the instantaneous velocity.

**In a circular motion, the instantaneous acceleration of the body is always perpendicular to the instantaneous velocity, and that acceleration is called centripetal acceleration. The speed remains unchanged; only the direction changes as the perpendicular acceleration changes the body’s trajectory.**