## Difference Between Speed And Velocity

- The ‘distance’ travelled by a body in unit time interval is called its
**speed**. When the position of a body changes in particular direction, then speed is denoted by ‘velocity’. i.e. the rate of change of displacement of a body is called its**Velocity**. - Speed is a scalar quantity while velocity is a vector quantity.
- \( \text{Speed}=\frac{\text{distance}}{\text{time}} \)
- \( \text{Velocity}=\frac{\text{displacement}}{\text{time}} \)
- Unit: In M.K.S. system = ms
^{-1}

In C.G.S. system = cm/s - If time distance graph is given then speed can be given by the slope of the line, at given time

\( \text{V}=\frac{\text{ }\!\!\Delta\!\!\text{ s}}{\text{ }\!\!\Delta\!\!\text{ t}}=\text{Slope} \)

- The area of velocity time graph gives displacement travelled.

**Types of speed**

**(a) Average and Instantaneous speed**

**Average speed: **It is obtained by dividing the total distance travelled by the total time interval. i.e.

\( \text{Average speed}=\frac{\text{total}\,\,\text{distance}}{\text{total}\,\,\text{time}} \)

\( \text{Average}\,\text{velocity}=\frac{\text{displacement}}{\text{total}\,\,\text{time}} \)

- Average speed is a scalar, while average velocity is a vector.
- For a moving body average speed can never be –ve or zero (unless t → ∞), while average velocity can be i.e

\( {{v}_{av~}}>0\text{ while }\overset{\to }{\mathop{{{v}_{av}}}}\,>=or<\text{ }0 \) - In general average speed is not equal to magnitude of average velocity. However it can be so if the motion is along a straight line without change in direction
- If a particle travels distances L
_{1}, L_{2}, L_{3}at speeds v_{1}, v_{2}, v_{3}etc respectively, then

\( {{v}_{av~}}=\frac{\Delta s}{\Delta t}=\frac{{{L}_{1}}+{{L}_{2}}+…..+{{L}_{n}}}{\frac{{{L}_{1}}}{{{v}_{1}}}+\frac{{{L}_{2}}}{{{v}_{2}}}+….+\frac{{{L}_{n}}}{{{v}_{n}}}}=\frac{\sum{Li}}{\sum{\frac{{{L}_{i}}}{{{v}_{i}}}}} \) - If a particle travels at speeds v
_{1}, v_{2}etc for intervals t_{1}, t_{2}etc respectively, then

\( {{v}_{av~}}=\frac{{{v}_{1}}{{t}_{1}}+{{v}_{2}}{{t}_{2}}+….}{{{t}_{1}}+{{t}_{2}}+….}=\frac{\sum{{{v}_{1}}{{t}_{1}}}}{\sum{{{t}_{1}}}} \)

**Instantaneous speed: **The speed of a body at a particular instant of time is called its instantaneous speed.

\( =\underset{\Delta t\to 0}{\mathop{\lim }}\,\,\frac{\Delta s}{\Delta t}=\frac{ds}{dt} \)

**(b) Uniform and Non uniform speed**

**Uniform speed: **If an object covers equal distance in equal interval of time, then time speed graph of an object is a straight line parallel to time axis then body is moving with a uniform speed.

**Non-uniform speed: **If the speed of a body is changing with respect to time it is moving with a non-uniform speed.

## Speed And Velocity Example Problems With Solutions

**Example 1. **The distance between two points A and B is 100 m. A person moves from A to B with a speed of 20 m/s and from B to A with a speed of 25 m/s. Calculate average speed and average velocity.

**Solution: (i)** Distance from A to B = 100 m

Distance from B to A = 100 m

Thus, total distance = 200 m

Time taken to move from A to B, is given by

\( {{t}_{1}}=\frac{\text{distance}}{\text{velocity}}=\frac{100}{20}=5\text{ seconds} \)

Time taken from B to A, is given by

\( {{t}_{2}}=\frac{\text{distance}}{\text{velocity}}=\frac{100}{25}=4\text{ seconds} \)

Total time taken = t_{1} + t_{2} = 5 + 4 = 9 sec.

∴ Average speed of the person

\( =\frac{\text{Total}\,\text{dis}\,\text{tan}\,\text{cecovered}}{\text{Total}\,\text{time}\,\text{taken}}=\frac{200}{9}=22.2\text{ m/s} \)

**(ii)** Since person comes back to initial position A, displacement will be zero, resulting zero average velocity.

**Example 2. **A car moves with a speed of 40 km/hr for first hour, then with a speed of 60 km/hr for next \(1\frac{1}{2}\) half hour and finally with a speed of 30 km/hr for next hours. Calculate the average speed of the car.

**Solution: **Distance travelled in first hour, is given by

s_{1} = speed × time = 40 km/hr × 1 hr = 40 km

Distance travelled in next half an hour, is given by

s_{2} = speed × time = 60 km/hr × \(\frac { 1 }{ 2 }\) hr = 30 km

Distance travelled in last \(1\frac{1}{2}\) hours, is given by

s_{3} = speed × time = 30 km/hr × \(\frac { 3 }{ 2 }\) hr = 45 km

Thus, total distance travelled = s_{1} + s_{2} + s_{3}

= 40 + 30 + 45 = 115 km

Total time taken = 1 + \(\frac { 1 }{ 2 }\) + \(1\frac{1}{2}\) = 3 hours

Average speed = \(\frac { Total distance covered }{ Total time taken } \) = \(\frac { 115km }{ 3hrs }\)

= 38.33 km/hr

**Example 3. **Figure shows time distance graph of an object. Calculate the following :

(i) Which part of the graph shows that the body is at rest ?

(ii) Average speed in first 10 s.

(iii) Speeds in different parts of motion.

**Solution: (i)** The part BC shows that the body is at rest.

**(ii)** In first 10 seconds, distance travelled = 100m

\( \text{Average speed}=\frac{\text{total}\,\,\text{distance}}{\text{total}\,\,\text{time}} \)

\( =\frac{100}{10}=10\text{ m/s} \)

**(iii)** Speed of the object in part AB is given by slope = 100/6 = \(\frac { 50 }{ 3 }\) m/s

Speed of object in part BC = 0 m/s

Speed of the object in part CD

\( =\frac{100-40}{12-10}=\frac{60}{2}=30~\text{m/s} \)

Speed of object in part DE

\( =\frac{40-0}{14-12}=\frac{40}{2}=30~\text{m/s} \)

**Example 4. **Time-velocity graph of a particle is shown in Figure. Calculate the distance travelled in first seconds.

**Solution: **Distance travelled in first 8s is given by area OABCG

= area of rectangle OAMG + area of triangle BMC

= 8 × 60 + \(\frac { 1 }{ 2 }\) × 4 × 40

= 480 + 80 = 560 m.

**Example 5. **A cow walked along a curved path from P to Q, which is 70 m away from P. Q lies to the south-west of P. The distance travelled by the cow is 240 m and the time taken is 160 s.

Calculate the

(a) average speed,

(b) average velocity,

of the cow moving from P to Q.

**Solution:**

Total distance travelled = 240 m

Displacement = 70 m

Time taken = 160 s