Difference Between Speed and Velocity

The rate of change of position of an object with time in any direction is called its speed. speed has only magnitude and no direction, so it is a scalar quantity. Different types of speeds exist. The speedometer of an automobile indicates its instantaneous speed at any instant.

The rate of change of position of an object with time in a given direction is called its velocity. Velocity has both magnitude and direction, so it is a vector quantity. Different types of velocity exist.

In this article, we will discuss speed, and types of speed, velocity and types of velocity, speed and velocity difference with numerical examples. Speed and velocity is an important topic in Class 9 physics, It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams.

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This Story also Contains

1. What is Speed?
2. Velocity
3. Solved Examples Based on Speed And Velocity

What is Speed?

Speed is the quantity which shows how fast the body is moving. It does not have any direction associated with it.

Formula to calculate speed:

$$\text{ Speed = Distance/Time }$$

Numerical example:

1- If a body covers a distance of 18m in 1 sec then, find the speed of the body.

The speed of a body is calculated using the formula:

$$\text{ Speed }=\frac{\text{ Distance }}{\text{ Time }}$$

Here, the distance covered is 18 meters, and the time taken is 1 second. Plugging in the values:

$$\text{ Speed }=\frac{18\mathrm{m}}{1\mathrm{s}}=18\mathrm{m}/\mathrm{s}$$

So, the speed of the body is $18\mathrm{m}/\mathrm{s}$.

After speed, now let's shift to the concept of average and instantaneous speed.

Average Speed and Instantaneous Speed

Average Speed: Amount of total distance covered in total time.

Mathematically average speed can written as,

Average Speed $=\frac{\text{ Total Distance Traveled }}{\text{ Total Time Taken }}$

Numerical example:

A body covers a total distance of 50 m with variable speed in 5 sec. Find the average speed.

$$\text{ Average Speed }=\frac{\text{ Total Distance }}{\text{ Total Time }}$$

Substituting these values:

$$\text{ Average Speed }=\frac{50\mathrm{m}}{5\mathrm{s}}=10\mathrm{m}/\mathrm{s}$$

Tips to Calculate Average Speed

The formula for average speed when a body covers two distances $s_1$ and $s_2$ in times $t_1$ and $t_2$ respectively is:

$$V_{\mathrm{avg}}=\frac{s_1+s_2}{t_1+t_2}$$

Instantaneous Speed:

It is the speed at that particular instant or small interval of time.

Mathematically,

$$\text{ Instantaneous Speed }=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{\mathrm{\Delta }s}{\mathrm{\Delta }t}$$

where:

• $\mathrm{\Delta }s$ is the displacement over a very small time interval $\mathrm{\Delta }t$,
• $\underset{\mathrm{\Delta }t\to 0}{lim}$ represents taking the limit as the time interval approaches zero.

Also read -

• NCERT Solutions for All Subjects
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Velocity

Velocity is the rate of change of displacement. It is the displacement in unit time. It is a vector quantity. The S.I. unit of measurement of velocity is ms^-1.

$$\text{ Velocity }=\frac{\text{ Displacement }}{\text{ Time }}$$

where:

• Displacement is the change in position of the object in a specific direction,
• Time is the time taken for the displacement to occur.

Now let's understand about the concept of average and instantaneous velocity.

Average Velocity and Instantaneous Velocity

Average Velocity: Amount of total displacement covered in total time.

$$\text{ Average Velocity }=\frac{\text{ Total Displacement }}{\text{ Total Time }}$$

If the displacement is represented by $\mathrm{\Delta }s$ and the total time by $\mathrm{\Delta }t$, then:

$$\text{ Average Velocity }=\frac{\mathrm{\Delta }s}{\mathrm{\Delta }t}$$

Instantaneous Velocity: It is Velocity at that particular instant or small interval of time.

Instantaneous Velocity $=\frac{ds}{dt}$
where:

• $\frac{ds}{dt}$ is the derivative of displacement $s$ with respect to time $t$,
• This derivative represents how displacement changes over an infinitesimally small time interval, giving the velocity at that exact moment.

Relation between Speed and Velocity:

Speed and velocity have many similaities and differences between them. But majorly direction is the main point of difference between them. Basically, speed do not contain direction but velocity conatins direction. Here we will discuss the similarities and difference between speed and velocity in detail. "Velocity is speed with displacement," asserts the relationship between speed and velocity.

Similarities between speed and velocity:

• Both speed and velocity are ways of calculating an object's change in position over time.
• In practice, an object's speed and velocity are the same in a straight line motion. Since distance and displacement will be the same.
• Because they are both physical quantities, they can both be measured and quantified.
• The units of speed and velocity are the same, metres per second or m/s.

Difference between speed and velocity:

Definition :

• Speed : It is the rate at which an object covers distance.
• Velocity : The rate at which an object changes its position (displacement) in a specific direction.

Type of Quantity :

• Speed : It is a Scalar quantity (only has magnitude, no direction)
• Velocity : It is a Vector quantity (has both magnitude and direction).

Formula :

• Speed: Speed $=\frac{\text{ Distance }}{\text{ Time }}$
• Velocity: Velocity $=\frac{\text{ Displacement }}{\text{ Time }}$

Direction :

• Speed : Does not indicate any direction.
• Velocity : It indicates the direction of motion.

Values :

• Speed : Always positive or zero. Can never be negative.
• Velocity : This can be positive, negative, or zero, depending on direction.

Example :

• Speed : "A car travels at 60 km/h."
• Velocity : "A car moves north at 60 km/h."

Above points will help the students to distinguish between the speed and the velocity.

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Solved Examples Based on Speed And Velocity

Example 1: A body travels 100 km southwest and then 502 km in the northern direction. The total magnitude of the velocity if the time taken is 1hr.

1) 100+502 km/hr
2) 100−502 km/hr
3) 502 km/hr
4) 100 km/hr

Solution:

As we learned,

Average Velocity = Total Displacement Total time taken

So,

Thus, the average velocity is calculated as:

$$\text{ Average Velocity }=\frac{\text{ Total Displacement }}{\text{ Total Time }}=\frac{502\mathrm{km}}{1\mathrm{hr}}=502\mathrm{km}/\mathrm{hr}$$

Example 2: An object moving with a speed of 6.25 ms−1, is decelerated at a rate given by dvdt=−2.5v where v is the instantaneous speed. The time taken (in seconds) by the object, to come to rest, would be:

1) 2

2) 1

3) 4

4) 8

Solution:

Step 1: Separate Variables
Rewrite the equation by separating $v$ and $t$ :

$$\frac{dv}{v}=-2.5dt$$

Step 2: Integrate Both Sides
Integrate both sides to find $v$ as a function of $t$ :

$$\int \frac{1}{v}dv=\int -2.5dt$$

This yields:

$$\ln v=-2.5t+C$$

Step 3: Solve for the Constant $C$
At $t=0$, the initial speed $v=6.25\mathrm{m}/\mathrm{s}$ :

$$\ln (6.25)=C$$
So, $C=\ln (6.25)$.

Step 4: Substitute $C$ and Solve for $t$ when $v=0$
Now, the equation becomes:

$$\ln v=-2.5t+\ln (6.25)$$

To solve for $t$ when $v\to 0$, take the limit of both sides as $v$ approaches zero:

$$-2.5t=-\ln (6.25)$$

Solving for $t$ :

$$t=\frac{\ln (6.25)}{2.5}\approx 2\text{ seconds }$$

Hence, the answer is option (1).

Example 3: A particle moves such that its position vector r^(t)=cos⁡ωti^+sin⁡ωtj^ where ω is a constant and t is time. Then which of the following statements is true for the velocity v→(t) and acceleration a→(t) of the particle :
1) v→ and a→ both are parallel to r→
2) v→ is perpendicular to r→ and a→ is directed away from the origin
3) v→ and a→ both are perpendicular to r→
4) v→ is perpendicular to r→ and a→ is directed towards the origin

Solution:

Given $\overrightarrow{r}(t)=\cos (\omega t)\hat{i}+\sin (\omega t)\hat{j}$
1. Velocity $\overrightarrow{v}(t)$ : Calculating $\overrightarrow{v}(t)$, we get $\overrightarrow{v}(t)=-\omega \sin (\omega t)\hat{i}+\omega \cos (\omega t)\hat{j}$, which is perpendicular to $\overrightarrow{r}(t)$ (dot product $=0$ ).
2. Acceleration $\overrightarrow{a}(t)$ : Calculating $\overrightarrow{a}(t)$, we get $\overrightarrow{a}(t)=-{\omega }^2\overrightarrow{r}(t)$, meaning $\overrightarrow{a}(t)$ is directed towards the origin.

Hence, the answer is the option (4).

Example 4: A particle moves 50 m in 5 seconds, then 20 m in the next 4 seconds and 30 m in the next 7 seconds, then the average speed (in m/s ) of the particle is :

1) 6.25

2) 7.25

3) 8.50

4) 5

Solution:

Step 1: Calculate Total Distance and Total Time
- Total Distance $=50+20+30=100\mathrm{m}$
- Total Time $=5+4+7=16\mathrm{s}$

Step 2: Calculate Average Speed

$$\text{ Average Speed }=\frac{\text{ Total Distance }}{\text{ Total Time }}=\frac{100}{16}=6.25\mathrm{m}/\mathrm{s}$$

Hence, the answer is the option (1).

Example 5: A particle travelled first 10 m with 2 m/s, the next 10 m with 3 m/s and the last 10 m with 6m/s then its average speed is (in m/s) :

1) 3

2) 4

3) 5

4) 6

Solution:

Total Distance $=10+10+10=30\mathrm{m}$

Total Time $=\frac{10}{2}+\frac{10}{3}+\frac{10}{6}=10$ seconds

Average Speed $=\frac{\text{ Total Distance }}{\text{ Total Time }}=\frac{30}{10}=3\mathrm{m}/\mathrm{s}$

Hence, the answer is the option (1).