How to Find Velocity Vector: A Complete Guide

Velocity vector is a fundamental concept in physics and mathematics that describes the motion of an object. It provides information about both the speed and direction of an object’s movement. In this blog post, we will explore various techniques to find the velocity vector, along with special cases and worked-out examples. So, let’s dive in!

Techniques to Find Velocity Vector

How to Find Velocity Vector from Position Vector and Speed

One of the ways to find the velocity vector is by using the position vector and speed of an object. The position vector represents the displacement of the object from a reference point, while the speed denotes the rate of change of position. To find the velocity vector, we can simply divide the position vector by the speed. Mathematically, it can be expressed as:

 vec{v} = frac{vec{r}}{t}

where  vec{v} is the velocity vector,  vec{r} is the position vector, and  t is the time taken.

Let’s consider an example to understand this better. Suppose a car is moving in a straight line, and its position vector is  vec{r} = 5hat{i} + 12hat{j}  hat{i} " class="ql-img-inline-formula quicklatex-auto-format" decoding="async" height="25" loading="lazy" data-src="https://techiescience.com/wp-content/ql-cache/quicklatex.com-4bc6024073e8e5f32cae5980217b1d92_l3.png" style="vertical-align: -5px;" title="Rendered by QuickLaTeX.com" width="124"/> and <img alt=" hat{j} " class="ql-img-inline-formula quicklatex-auto-format" decoding="async" height="24" loading="lazy" data-src="https://techiescience.com/wp-content/ql-cache/quicklatex.com-d312a1b35277d4e8a742a254b6961dc6_l3.png" style="vertical-align: -4px;" title="Rendered by QuickLaTeX.com" width="10"/> are unit vectors in the x and y directions, respectively and the speed is 2 m/s. We can find the velocity vector as follows:

 vec{v} = frac{5hat{i} + 12hat{j}}{2} = 2.5hat{i} + 6hat{j}

Hence, the velocity vector of the car is 2.5 m/s in the x-direction and 6 m/s in the y-direction.

How to Find Velocity Vector from Parametric Equations

In some cases, the motion of an object can be described using parametric equations, where the position of the object is given as a function of time. To find the velocity vector from parametric equations, we need to differentiate the position equations with respect to time. Let’s consider a particle moving in a plane with parametric equations:

 x = f(t)

 y = g(t)

The velocity vector can be found by differentiating both equations with respect to time:

 xhat{i} + yhat{j}" class="ql-img-inline-formula quicklatex-auto-format" decoding="async" height="27" loading="lazy" data-src="https://techiescience.com/wp-content/ql-cache/quicklatex.com-2541e78ad8693bc20ea46677147605af_l3.png" style="vertical-align: -7px;" title="Rendered by QuickLaTeX.com" width="182"/> = frac{dx}{dt}hat{i} + frac{dy}{dt}hat{j}

This gives us the velocity vector in terms of the derivatives of the position equations.

How to Find Velocity Vector from Acceleration Vector

Acceleration is the rate of change of velocity, and it plays a crucial role in determining the velocity vector. If we have the acceleration vector, we can integrate it with respect to time to find the velocity vector. Mathematically, it can be written as:

 vec{v} = int vec{a} dt

where  vec{v} is the velocity vector and  vec{a} is the acceleration vector.

How to Find Velocity Vector Given Position Vector

If we are given the position vector of an object as a function of time, we can find the velocity vector by differentiating the position vector with respect to time. Mathematically, it can be expressed as:

 vec{v} = frac{dvec{r}}{dt}

where  vec{r} is the position vector and  t is the time variable.

How to Find Velocity Vector Given Speed and Angle

In certain cases, we may be given the speed of an object and the angle at which it is moving. To find the velocity vector in such cases, we can use trigonometric functions to determine the components of the velocity vector in the x and y directions. Let’s say the speed is  V and the angle is  theta . Using trigonometry, we can find the x and y components of the velocity vector as:

 theta" class="ql-img-inline-formula quicklatex-auto-format" decoding="async" height="21" loading="lazy" data-src="https://techiescience.com/wp-content/ql-cache/quicklatex.com-e8ca3a07b1cc65654fa1486cf7a31d7b_l3.png" style="vertical-align: -5px;" title="Rendered by QuickLaTeX.com" width="168"/>

 theta" class="ql-img-inline-formula quicklatex-auto-format" decoding="async" height="22" loading="lazy" data-src="https://techiescience.com/wp-content/ql-cache/quicklatex.com-7ac25a01f29a15cfc8bffc8bf538414b_l3.png" style="vertical-align: -6px;" title="Rendered by QuickLaTeX.com" width="165"/>

The velocity vector can then be expressed as  vec{v} = V_x hat{i} + V_y hat{j} .

Special Cases in Finding Velocity Vector

How to Find Velocity Vector in Projectile Motion

how to find velocity vector
Image by Maschen – Wikimedia Commons, Wikimedia Commons, Licensed under CC0.

Projectile motion refers to the motion of an object that is launched into the air and follows a curved path due to the influence of gravity. To find the velocity vector in projectile motion, we can break it down into its horizontal and vertical components. The horizontal component remains constant throughout the motion, while the vertical component changes due to gravity. By combining the horizontal and vertical components, we can determine the velocity vector at any given point in time.

How to Find Initial Velocity Vector

how to find velocity vector
Image by Avdimetaj – Wikimedia Commons, Wikimedia Commons, Licensed under CC BY-SA 3.0.
velocity vector 3

The initial velocity vector represents the velocity of an object at the start of its motion. It can be found using the same techniques mentioned earlier, such as using the position vector and speed, parametric equations, or differentiating the position vector with respect to time.

How to Find Resultant Velocity Vector

The resultant velocity vector is the vector sum of two or more velocity vectors. It can be found using vector addition, where the individual velocity vectors are added together to obtain the resultant. Mathematically, it can be expressed as:

 vec{v}_{text{resultant}} = vec{v}_1 + vec{v}_2 + ldots + vec{v}_n

where  vec{v}_1, vec{v}_2, ldots, vec{v}_n are the individual velocity vectors.

How to Find Velocity Vector at Maximum Height

When an object is launched vertically upwards and reaches its maximum height, its velocity vector at that point can be determined. At the maximum height, the object momentarily comes to rest before coming back down. The velocity vector at the maximum height will have a magnitude of zero, as the object has zero speed. However, it will still have a direction, which depends on the initial velocity and the direction of motion.

How to Find Final Velocity Vector

The final velocity vector represents the velocity of an object at the end of its motion. It can be found using similar techniques as those used for finding the initial velocity vector, such as using the position vector and speed, parametric equations, or differentiating the position vector with respect to time.

How to Find Angular Velocity Vector

In rotational motion, the angular velocity vector describes the rate of change of angular displacement of an object. It is perpendicular to the plane of rotation and has a magnitude proportional to the angular speed. The direction of the angular velocity vector follows the right-hand rule, where the fingers of the right hand curl in the direction of rotation, and the thumb points in the direction of the angular velocity vector.

Worked Out Examples

Let’s now work through some examples to solidify our understanding of finding velocity vectors.

Example of Finding Velocity Vector from Position Vector and Speed

Suppose a particle has a position vector  vec{r} = 4hat{i} + 3hat{j} and a speed of 5 m/s. To find the velocity vector, we divide the position vector by the speed:

 vec{v} = frac{4hat{i} + 3hat{j}}{5} = 0.8hat{i} + 0.6hat{j}

Thus, the velocity vector of the particle is 0.8 m/s in the x-direction and 0.6 m/s in the y-direction.

Example of Finding Velocity Vector from Parametric Equations

velocity vector 2

Consider a particle moving in a plane with the following parametric equations:

 x = 2t^3

 y = 3t^2

To find the velocity vector, we differentiate both equations with respect to time:

 frac{dx}{dt} = 6t^2

 frac{dy}{dt} = 6t

Thus, the velocity vector is given by  vec{v} = 6t^2hat{i} + 6that{j} .

Example of Finding Resultant Velocity Vector

Suppose a car is moving with a velocity of 20 m/s in the x-direction and 15 m/s in the y-direction. Another car is moving with a velocity of 10 m/s in the x-direction and 5 m/s in the y-direction. To find the resultant velocity vector, we add the individual velocity vectors:

 20hat{i} + 15hat{j}" class="ql-img-inline-formula quicklatex-auto-format" decoding="async" height="25" loading="lazy" data-src="https://techiescience.com/wp-content/ql-cache/quicklatex.com-100c2a2f6d5e90916b3d8410d5ff64d5_l3.png" style="vertical-align: -5px;" title="Rendered by QuickLaTeX.com" width="247"/> + <img alt=" 10hat{i} + 5hat{j}" class="ql-img-inline-formula quicklatex-auto-format" decoding="async" height="24" loading="lazy" data-src="https://techiescience.com/wp-content/ql-cache/quicklatex.com-6fa233890cb47612806075bbf23808a7_l3.png" style="vertical-align: -4px;" title="Rendered by QuickLaTeX.com" width="73"/> = 30hat{i} + 20hat{j}

Hence, the resultant velocity vector is 30 m/s in the x-direction and 20 m/s in the y-direction.

In this blog post, we have explored various techniques to find the velocity vector, including using position vectors and speed, parametric equations, acceleration vectors, and differentiation of position vectors. We have also discussed special cases such as projectile motion, initial velocity vector, resultant velocity vector, and velocity vectors in different scenarios. By understanding and applying these techniques, we can accurately determine the velocity vectors of objects in motion. Keep practicing and experimenting with different examples to strengthen your grasp of this important concept in physics and mathematics. Happy learning!

Numerical Problems on how to find velocity vector

Problem 1:

A car is moving in a straight line with a constant acceleration of . The initial velocity of the car is and the initial position is . Find the velocity vector of the car at a given time .

Solution:

Given:
Acceleration ((a)) =  2 , text{m/s}^2
Initial velocity ((v_0)) =  5 , text{m/s}
Initial position ((x_0)) =  3 , text{m}
Time ((t)) =  4 , text{s}

We know that the velocity vector  (vec{v}) can be calculated using the equation:

 vec{v} = vec{v_0} + vec{a}t

First, let’s calculate the change in velocity using the formula:

 Delta vec{v} = vec{a}t

Substituting the given values, we get:

 Delta vec{v} = 2 , text{m/s}^2 times 4 , text{s} = 8 , text{m/s}

Now, we can calculate the velocity vector  (vec{v}) by adding the initial velocity  (vec{v_0}) and the change in velocity  (Delta vec{v}):

 vec{v} = vec{v_0} + Delta vec{v}

Substituting the given values, we get:

 vec{v} = 5 , text{m/s} + 8 , text{m/s} = 13 , text{m/s}

Therefore, the velocity vector of the car at  t = 4 , text{s} is  13 , text{m/s}.

Problem 2:

velocity vector 1

A particle is moving in a two-dimensional plane with a constant acceleration given by  vec{a} = 3hat{i} + 4hat{j} , text{m/s}^2. At (t = 0), the particle has an initial velocity of  vec{v_0} = 2hat{i} - hat{j} , text{m/s} and is located at the origin ((x = 0), (y = 0)). Find the velocity vector of the particle at a given time  t = 2 , text{s}.

Solution:

Given:
Acceleration  (vec{a}) =  3hat{i} + 4hat{j} , text{m/s}^2
Initial velocity  (vec{v_0}) =  2hat{i} - hat{j} , text{m/s}
Time ((t)) =  2 , text{s}

To find the velocity vector  (vec{v}), we can use the equation:

 vec{v} = vec{v_0} + vec{a}t

Multiplying the acceleration vector by time, we have:

 3hat{i} + 4hat{j}" class="ql-img-inline-formula quicklatex-auto-format" decoding="async" height="25" loading="lazy" data-src="https://techiescience.com/wp-content/ql-cache/quicklatex.com-e7902fa96d0e8f760ccc6ed283923893_l3.png" style="vertical-align: -5px;" title="Rendered by QuickLaTeX.com" width="169"/> times 2 , text{s} = 6hat{i} + 8hat{j} , text{m/s}<br/>

Now, we can calculate the velocity vector  (vec{v}) by adding the initial velocity  (vec{v_0}) and the product of acceleration and time:

 vec{v} = vec{v_0} + vec{a}t

Substituting the given values, we get:

 2hat{i} - hat{j}" class="ql-img-inline-formula quicklatex-auto-format" decoding="async" height="25" loading="lazy" data-src="https://techiescience.com/wp-content/ql-cache/quicklatex.com-03e3f82936e9e1329ac127297a57257f_l3.png" style="vertical-align: -5px;" title="Rendered by QuickLaTeX.com" width="151"/> + <img alt=" 6hat{i} + 8hat{j}" class="ql-img-inline-formula quicklatex-auto-format" decoding="async" height="24" loading="lazy" data-src="https://techiescience.com/wp-content/ql-cache/quicklatex.com-4b7a4dc9d042179b742d953aff479ce1_l3.png" style="vertical-align: -4px;" title="Rendered by QuickLaTeX.com" width="64"/> = 8hat{i} + 7hat{j} , text{m/s}<br/>

Therefore, the velocity vector of the particle at  t = 2 , text{s} is  8hat{i} + 7hat{j} , text{m/s}.

Problem 3:

A rocket is launched vertically upwards with an initial velocity of  20 , text{m/s} and experiences a constant acceleration due to gravity of  9.8 , text{m/s}^2. Find the velocity vector of the rocket after  5 , text{s} from the time of launch.

Solution:

Given:
Initial velocity ((v_0)) =  20 , text{m/s}
Acceleration due to gravity ((g)) =  9.8 , text{m/s}^2
Time ((t)) =  5 , text{s}

Since the rocket is moving vertically upwards, the acceleration due to gravity will act in the opposite direction to the motion. Therefore, we can take the acceleration as negative.

To find the velocity vector ((v)), we can use the equation:

 v = v_0 + gt

Substituting the given values, we get:

 -9.8 , text{m/s}^2" class="ql-img-inline-formula quicklatex-auto-format" decoding="async" height="27" loading="lazy" data-src="https://techiescience.com/wp-content/ql-cache/quicklatex.com-ef29a10055d6ff9a20d14a919ffbad99_l3.png" style="vertical-align: -6px;" title="Rendered by QuickLaTeX.com" width="285"/> times 5 , text{s} = 20 , text{m/s} - 49 , text{m/s} = -29 , text{m/s}<br/>

Therefore, the velocity vector of the rocket after  5 , text{s} from the time of launch is  -29 , text{m/s}.

Also Read: