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# Speed and Velocity – Description with practice problems

By going through this article, you will get a deep knowledge about the definitions, equations, classifications and differences between speed and velocity, steps to calculate the speed and velocity of a moving body with the help of solved problems.

## What does speed mean and what is its formula?

Speed of a particle is defined as the distance covered by it in unit time or it is the time rate of change of distance. large begin{aligned}i.e., Speed = frac{Distance;travelled }{Time}end{aligned}

The SI unit of speed is metre/second and its dimensional formula is [M0L1T-1]

Speed gives only a quantitative idea about how fast or how slow an object is moving. The direction of motion is not considered. Hence speed is a scalar quantity and it can never be a negative value.

### Different types of speed

Basically, there are four different types of speed in physics and each are explained below:

1. #### Uniform Speed:

If a body covers equal distances in equal intervals of time, it is said to be moving with uniform speed.

2. #### Variable Speed:

If the body covers equal distances in unequal intervals of time, it is said to be moving with variable speed.

3. #### Instantaneous Speed:

The speed of an object at a particular instant of time is called its instantaneous speed.

4. #### Average Speed:

The ratio of total distance travelled by a body to the total time taken is called average speed. large begin{aligned}i.e., Average;Speed = frac{Total;distance;travelled }{Total;time;taken}end{aligned}

• If a particle travels with a speed s1, for a time interval t1 and with a speed s2 for a time interval t2,
then , total distance = s1t1 + s2t2
total time = t1 + t2 large begin{aligned} therefore Average;Speed,;S_{avg} &= frac{s_{1}t_{1} + s_{2}t_{2} }{t_{1} + t_{2} } \If;t_{1} = t_{2}=t,;S_{avg} &=frac{s_{1}+s_{2}}{2} end{aligned}
• If a particle covers a distance d1 with a speed s1 and then d2 with s2 ,
then, total distance = d1 + d2 large begin{aligned} Also, ;total ; time &= frac{d_{1}}{s_{1}}+frac{d_{2}}{s_{2}} \therefore Average;Speed,;S_{avg} &= frac{d_{1} + d_{2}}{left ( frac{d_{1}}{s_{1}}+frac{d_{2}}{s_{2}} right )} \If;d_{1} = d_{2}=d,;S_{avg} &=frac{2s_{1}s_{2}}{s_{1}+s_{2}} end{aligned}

## What does velocity mean and what is the formula to find velocity?

The velocity of a moving particle is the distance travelled by it in unit time in a given direction, or in other words, velocity is the rate of displacement of a body. Large begin{aligned} i.e.,; Velocity,; v = frac{Displacement }{Time;taken} end{aligned}

If x1 and x2 are the position coordinates of a particle in uniform motion at the instants t1 and t2 respectively, its velocity is given by,  Large begin{aligned} v = frac{x_{2} - x_{1} }{t_{2} - t_{1}} end{aligned}

The SI unit of velocity is metre/second and its dimensional formula is [M0L1T-1]

The velocity is a vector quantity. It can have positive, negative or zero value. It is positive when the displacement is positive(i.e., when the particle moves in the direction of increasing position co-ordinate) and negative when the displacement is negative.

### Types of velocity

The different types of velocities are uniform velocity, variable velocity, average velocity and instantaneous velocity. Now let us discuss each one in detail.

1. #### Uniform velocity

A particle is said to be moving with uniform velocity if it covers equal displacements in equal intervals of time, that is, velocity remains constant throughout the motion. That is, when a body is moving with uniform velocity, the magnitude and direction of velocity remain the same at any point.
In this case, the displacement ‘s’ of the particle during a given time interval ‘t’ is obtained by taking the product of the velocity ‘v’ and the time interval ‘t’.
i.e., Displacement = velocity × time interval
or, s = vt

2. #### Variable velocity or non-uniform velocity

A body is said to be moving with variable velocity if its speed or direction or both change with time.
For example, consider a body circulating around a circle with uniform speed. Its velocity is variable as its direction changes with motion.

3. #### Average velocity

Consider a body moving with variable velocity. Then the ratio of total displacement to the total time interval is called average velocity. large begin{aligned} i.e.,;Average;velocity,;v_{av} &= frac{Total; Displacement}{Total;time;taken} \therefore Displacement &= average;velocity times time \ or, s &= v_{av}times t end{aligned}
That is, we can say average velocity of an object  during a given interval of time is the ratio of its displacement during that interval to the time taken.
Consider an object moving along the X – direction. Let x(t1) and x(t2) be its position co-ordinates at times t1 and t2 respectively. Then, large begin{aligned} Displacement &= xleft ( t_{2} right )-xleft ( t_{1} right ) \ Time;taken &= t_{2}-t_{1} \therefore Average;velocity,;v_{av} &= frac{xleft ( t_{2} right )-xleft ( t_{1} right )}{t_{2}-t_{1}} end{aligned}

4. #### Instantaneous velocity

Consider a body moving with variable velocity. The velocity of an object at a particular instant of time is called its instantaneous velocity.
For an object in motion along the X –direction, let x(t) and x(t + Δt) be the position co-ordinates at times t  and (t + Δt) respectively. Then its average velocity during the time interval t is, large begin{aligned} v_{avg} &= frac{xleft ( t+Delta t right )-xleft ( t right )}{left ( t+Delta t right )-t} \&= frac{Delta x}{Delta t} end{aligned}
Then the instantaneous velocity is given by large begin{aligned}V_{inst},;overline{v} &= lim_{Delta trightarrow 0}left ( frac{Delta x}{Delta t } right ) \&=frac{dx}{dt} end{aligned}
where Δx be the displacement of a body in an infinitely small interval of time Δt.
Here the time Δt tends to zero in the limit. The direction of instantaneous velocity is along the tangent to the path at the point.
Also, largebegin{aligned} frac{dx}{dt}end{aligned} is the derivative of x with respect to time t. In other words, the velocity of a particle at any instant is the derivative of the position co-ordinate of the particle with respect to time.

## Write the difference between speed and velocity in tabular form.

 Speed Velocity Speed is the rate of change of distance of a particle Velocity is the rate of displacement Speed is a scalar quantity Velocity is a vector quantity Speed can have only positive values Velocity can have positive, negative or zero values. Speed tells nothing about the direction of motion of the body Velocity gives the direction of motion of the body Its SI unit is m/s and has the dimension [M0L1T-1] Its unit and dimension is the same as that of speed which is m/s and [M0L1T-1]

## Speed and Velocity Problems with Solutions

Problem 1:
A motorcycle travels with a speed of 40 km/hr for 20 minutes and then with 50 km/hr for 5 minutes. Find the total distance travelled and average speed of the car.

Solution:
Given v1 = 40 km/hr
t1 = 20 min = 20/60 =  0.333 hr
v2 = 50 km/hr
t2 = 5 min = 5/60 = 0.083 hr

Distance travelled at the speed of 40 km/hr, d1 = v1t1 = 40 × 0.333 = 13.32 km
Distance travelled at the speed of 50 km/hr, d2 = v2t2 = 50 × 0.083 = 4.15 km
∴ Total distance travelled, D = d1+ d2  = 13.32 + 4.15 =  17.47 km

Average speed = Total distance / Total time = D / (t1+ t2 )
i.e., Savg   = 17.47 / (0.333 + 0.083) =  41.995 km/hr

Problem 2:
A bus covers the first half of the distance between two places at a speed of 70 km/hr and the second half at 40 km/hr. What is the average speed of the car?

Given s1 = 70 km/hr
s2 = 40 km/hr large begin{aligned} Average;speed,;s_{avg} &= frac{2s_{1}s_{2}}{s_{1}+s_{2}} \&= frac{2times 70times 40}{70+40} \&= 50.909;km/hr end{aligned}

Problem 3:
A vehicle covers 1/3 part of the total distance with a speed of 15 km/hr, second 1/3 part with a speed of 25 km/hr and the last 1/3 part with 50 km/hr. Find the average speed of the vehicle.

Solution:
Given s1 = 15 km/hr
s2 = 25 km/hr
s3 = 50 km/hr

Let D be the total distance covered.

Then the time to cover the first D/3 distance is, large begin{aligned} time,;t_{1} &= frac{distance}{speed} \&= frac{frac{D}{3}}{15} \&= frac{D}{3times 15} \&= frac{D}{45};hr end{aligned}

Similarly, time to cover the second D/3 distance is, large begin{aligned} time,;t_{2} &= frac{distance}{speed} \&= frac{frac{D}{3}}{25} \&= frac{D}{3times 25} \&= frac{D}{75};hr end{aligned}

Similarly, time to cover the third D/3 distance is, large begin{aligned} time,;t_{3} &= frac{distance}{speed} \&= frac{frac{D}{3}}{50} \&= frac{D}{3times 50} \&= frac{D}{150};hr end{aligned}

Therefore, the average speed is , large begin{aligned} therefore Average;speed,;s_{avg} &= frac{Total;distance}{Total;time} \&= frac{D}{frac{D}{45}+frac{D}{75}+frac{D}{150}} \&= frac{D}{Dleft [frac{1}{45}+frac{1}{75}+frac{1}{150}right ]} \&= frac{1}{left [frac{1}{45}+frac{1}{75}+frac{1}{150}right ]} \&= 23.684;km/hr end{aligned}

Problem 4
An object moves in a straight line along the X-axis, its displacement from the origin is given by the equation x = 5t3 – 4t2 + 2t, where x is in ‘metre’ and t in ‘second’. Find the average velocity of the body in the time interval from t = 0 to t = 2 seconds.

Given, x = 5t3 – 4t2 + 2t
Let x0   denotes the initial position.

When t = 0, x0   = 5×0 – 4×0 +2×0 = 0 (x0 denotes the initial position)
When t = 2, x1   = 5×23 – 4×22 +2×2 = 40-16+4 = 28 m
∴ Total displacement = x1   – x0  = 28-0 = 28 m
Total time = 2 second.

∴ Average velocity = Total displacement/Total time = 28/2 = 14 m/s

Problem 5
A boy walks for 1 minute at a speed of 1 m/s and then runs for 2 minutes at a speed of 4 m/s along a straight road. What is the average speed of the boy?

Given v1 = 1 m/s
t1 = 1 min = 60 s
v2 = 4 m/s
t2 = 2 min = 2 × 60  = 120 s

Distance travelled by boy at the speed of 1 m/s, d1 = v1t1 = 1 × 60 = 60 m
Distance travelled at the speed of 4 m/s, d2 = v2t2 = 4 × 120 = 480 m
∴ Total distance travelled, D = d1+ d2
= 60 + 480
=  540 m

Average speed, Savg = Total distance / Total time
= D / (t1+ t2 )
i.e., Savg   = 540 / (60 + 120)
=  3 m/s

Problem 6
A bus is moving at a speed of 55 km/hr. The bus has to cover a distance of 85 kilometres. What will be the time required by the bus to reach the destination?

Solution:
Given speed, s = 55 km/hr
Distance, d = 85 km
time, t = ?

We know, speed = distance / time
∴ time = distance / speed
= 85/55
=1.545 hr

i.e., the bus requires 1.545 hours to reach the destination.

Problem 7
A vehicle’s is described by the function x = 8t2 + 20t + 12. Calculate the instantaneous velocity of the vehicle at time t = 8 seconds.

Given the function, x = 8t2 + 20t + 12

The velocity at any instant of time can be find out by differentiating the provided function with respect to time. large begin{aligned} i.e., V_{inst},;overline{v}&=frac{dx}{dt} \&= frac{d}{dt}left ( 8t^{2}+20t+12 right ) \&= 8times 2t+20+0 \&= 16t + 20 end{aligned}

i.e., the instantaneous velocity at time, t = 8 seconds can be calculated as,

v(t) = 16t + 20
∴ At  t = 8 s,
v(8) = 16 × 8 + 20

= 128 + 20
= 148 m/s

i.e., the instantaneous veloity of the vehicle at time 8 seconds is 148 m/s.

Problem 8
A bus moves 98 km in 3 hours and 100 km in 2 hours in the same direction. a)Find the average speed of the bus for the whole journey. b) Find the average velocity of the bus for the whole journey.

Solution: a) Here, Total distance covered = 98 + 100 = 198 km
Total time taken = 3 + 2 = 5 hours
∴ The average speed of the bus for the whole journey = Total distance / Total time taken
= 198/5
= 39.6 km/hr

b) Given the whole journey is in the same direction.
∴ Total displacement =  98 + 100 = 198 km
Total time taken = 3 + 2 = 5 hours
∴ The average velocity of the bus for the whole journey = Total displacement/Total time taken
= 198/5
= 39.6 km/hr

Problem 9
A bus moves 90 km east in 3 hours and 50 km west in 2 hours. a)Find the average speed of the bus for the whole journey. b) Find the average velocity of the bus for the whole journey. a) Here, Total distance covered = 90 + 50 = 140 km
Total time taken = 3 + 2 = 5 hours
∴ The average speed of the bus for the whole journey = Total distance / Total time taken
= 140/5
= 28 km/hr

b) Given the whole journey is in opposite directions.
∴ Total displacement =  90 – 50 = 40 km
Total time taken = 3 + 2 = 5 hours
∴ The average velocity of the bus for the whole journey = Total displacement / Total time taken
= 40/5
= 8 km/hr

Problem 10
A boy starts running from a point on a circular path of a radius of 400 metres and comes back to the same point after 30 minutes. a)Find out the average speed of the running boy. b)Find out the average velocity of the running boy.

Solution:
a) The boy runs from a point in a circular path and comes back to the same point means he has covered a distance equal to the circumference of the circle.

Here, the radius of the circular path, r = 400 m
Time taken = 30 minutes
= 30 × 60 seconds
= 1800 s
∴ Distance covered = circumference of the circle
= 2πr
= 2 × 3.14 × 400
= 2512 metres

∴ The average speed of the running boy = Total distance/Total time taken
= 2512/1800
= 1.395 m/s

b) The boy runs from a point in a circular path and comes back to the same point means the displacement is equal to zero.

∴ The average velocity of the running boy = Total displacement/Total time taken
= 0/1800
= 0 m/s

Problem 11
A car moves north 160 km at 70 km/hr and then west 130 km at 80 km/hr.
a) Find the average speed of the car.

b)  Find the average velocity of the car. We know,
speed = distance/time

∴ time = distance / speed

a) Let t1 be the time to cover 160 km at a speed of 70 km/hr and is given by,
t1 = 160/70
= 2.286 hrs

Similarly, let t2 be the time to cover 130 km at a speed of 80 km/hr and is given by,
t2 = 130/80
= 1.625 hrs

∴ The average speed of the car = Total distance/Total time taken
= (160 + 130) / (2.286 + 1.625)
= 290 / 3.911
= 74.149 km/hr

b) Displacement is the shortest distance AC between the initial pont and final point and is calculated as follows.

By Pythagora’s theorem, AC2 = AB2 + BC2
i.e.,  AC = √(AB2 + BC2)
= √(1602 + 1302)
= √(25600 + 16900)
= √(42500)
= 206.155

∴  Average velocity of the car = Total displacement / Total time taken
= 206.155 / (t1 + t2)
= 206.155 / (2.286 + 1.625)
= 206.155 / 3.911
= 52.711 km/hr

Problem 12
The average speed of a car is 75 km/hr. How many kilometres the car will cover in 4 hours.

We know, Average speed = Total distance/Total time taken
∴ Total distance = Average speed × Total time taken

Given, average speed = 75 km/hr
Time taken = 4 hours
∴ Total distance = Average speed × Total time taken
= 75 × 4
= 300 km

This article explored the definitions, types and differences between speed and velocity of a moving object. Now you are able to calculate the speed and velocity of a moving body.