## What is Speed?

**Definition of Speed:** Speed is the rate of change of distance with time. Speed is different from velocity because itâ€™s not in a specified direction. In this article, you will learn how to solve speed, velocity, and acceleration problems.

Additionally, you need to know that **speed is a scalar quantity** and we can write its symbol as S. The formula for calculating the speed of an object is:

Speed, S = Distance (d) / Time (t)

Thus, s = d/t

**Note:** In most cases, we also use S as a symbol for distance.

The S.I unit for speed is meter per second (m/s) or ms^{-1}

**Non-Uniform or Average Speed:** This is a non-steady distance covered by an object at a particular period of time. We can also define non-uniform speed as the type of distance that an object covered at an equal interval of time.

The formula for calculating non-uniform speed is

**Average speed = Total distance covered by the object / Total time taken**

**Actual speed:** This is also known as the instantaneous speed of an object which is the distance covered by an object over a short interval of time.

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## What is Velocity?

**Definition of Velocity:** Velocity is the rate of displacement with time. Velocity is the speed of an object in a specified direction. The unit of velocity is the same as that of speed which is meter per second (ms^{-1}). We use V as a symbol for velocity.

**Note:** We often use U to indicate initial speed, and V to indicate final speed.

The formula for calculating **Velocity (V) = displacement (S) / time** **(t)**

The difference between **velocity and speed is the presence of displacement and distance** respectively. Because displacement is a measure of separation in a specified direction, while distance is not in a specified direction. Velocity is a vector quantity.

## Uniform Velocity

**Definition of uniform velocity:** The rate of change of displacement is constant no matter how small the time interval may be. Also, uniform velocity is the distance covered by an object in a specified direction in an equal time interval.

The formula for uniform velocity = Total displacement / Total time taken

## What is Acceleration?

**Definition of Acceleration:** Acceleration is the rate of change of velocity with time. Acceleration is measured in meters per second square (ms^{-2}). The symbol for acceleration is a. Acceleration is also a vector quantity.

The formula for acceleration, a = change in velocity (v)/time taken (t)

Thus, a = v/t

We can also write acceleration as

a = change in velocity/time = Î”V/Î”t = (v â€“ u)/t

[where v = final velocity, u = initial velocity, and t = time taken]

## Uniform Acceleration

In the case of uniform acceleration, the rate of change of velocity with time is constant.

Average velocity of the object = (Initial velocity + final velocity)/2

or

Average velocity = (v + u)/2

## What is Retardation?

Retardation is the decreasing rate of change in velocity moved, covered, or traveled by an object.

The formula for calculating retardation is

Retardation = Change in a decrease in velocity/time taken

## Equations of Motion

You can also apply the following equations of motion to calculate the speed, velocity, and acceleration of the body:

- v = u + at
- s = [ (v + u)/2 ]t
- v
^{2}= u^{2}+ 2as - s = ut + (1/2)at
^{2}

Where v = final velocity, u = initial velocity, t = time, a = acceleration, and s = distance

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## Solved Problems of Speed, Velocity, and Acceleration

Here are solved problems to help you understand how to calculate speed, velocity, and acceleration:

### Problem 1

A train moves at a speed of 54km/h for a one-quarter minute. Find the distance travelled by train.

**Solution**

**Data**

From the question above

Speed = 54 km/h = (54 x 1000)/(60 x 60) = 54,000/3,600 = 15 m/s

Time = one quarter minute = 1/4 minute = (1/4) x 60 = 15 seconds

Since we have

speed = distance/time

After cross-multiplication, we will now have

Distance = speed x time

We can now insert our data into the above expression

Distance = 15 m/s x 15 s = 225 m

Therefore, the distance travelled by train is 225 meters.

### Problem 2

A car travelled a distance of 5km in 50 seconds. Find the speed in meters per second.

**Solution**

**Data:**

Distance = 5km = 5 x 1000m = 5,000m

Time = 50 seconds

speed = ?

and the formula for speed

speed = distance/time = 5000/50 = 100m/s

### Problem 3

A motorcycle starting from rest moves with a uniform acceleration until it attains a speed of 108 kilometres per hour after 15 seconds. Find its acceleration.

**Solution**

**Data:**

From the question above

Initial velocity, u = 0 (because the motorcycle starts from rest)

Final velocity, v = speed = 108 km/h = (108 x 1000m) / (60 x 60s) = 108,000/3,600 = 30m/s

Time taken, t = 15 seconds

Therefore, we can now apply the formula that says

Acceleration = change in velocity/ time = (v-u)/t = (30-0)/15 =30/15 = 2ms^{-2}

Therefore, the acceleration of the motorcycle is 2ms^{-2}

### Problem 4

A bus covers 50 kilometres in 1 hour. What is it is the average speed?

**Solution**

**Data**

Total distance covered = 50 km = 50 x 1000m = 50,000m

Time taken = 1 hour = 1 x 60 x 60s =3,600s

Average speed = Total distance covered by the object / Total time taken

Therefore, we can now calculate the average speed of the bus by substituting our data into the above formula

Hence,

Average speed = 50,000/3,600 = 13.9 m/s

**Therefore, the average speed of the bus is 13.9m/s or approximately 14 meters per second.**

### Problem 5

A car travels 80 km in 1 hour and then another 20 km in the next hour. Find the average velocity of the car.

**Solution**

**Data:**

Initial displacement of the car = 80km

Final displacement of the car = 20km

The total displacement of the car = initial displacement of the car + final displacement of the car

Therefore,

The total displacement of the car = 80km + 20km = 100km

Also,

The time for 80km is 1hr

And the time for 20km is 1hr

Total time taken = The time for 80km (1hr) + The time for 20km (1hr)

Hence

Total time taken = 1hr + 1hr = 2hrs

Now, to calculate the average velocity of the car, we apply the formula that says

Average velocity = total displacement/total time taken = 100km/2hrs = 50km/h

We can further convert the above answer into meters per second

Average velocity = 50km/h = (50 x 1000m)/(1 x 60 x 60s) = 50,000/3,600 = 13.9m/s or 14ms^{-1}

**Therefore, the average velocity of the car is 14ms ^{-1}**

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### Problem 6

A body falls from the top of a tower 100 meters high and hits the ground in 5 seconds. Find its acceleration.

**Solution**

**Data**

Displacement = 100m

Time = 5 seconds

and we can apply the formula for acceleration that says

acceleration, a = Displacement/time^{2} = 100/5^{2} = 100/25 = 4ms^{-2}

**Therefore the acceleration due to the gravity of the body is 4ms ^{-2}**

**Note:** The acceleration due to the gravity of a body on the surface of the earth is constant (9.8ms^{-2}), even though there may be a slight difference due to the mass and altitude of the body.

### Problem 7

An object is thrown vertically upward at an initial velocity of 10ms^{-1} and reaches a maximum height of 50 meters. Find its initial upward acceleration.

**Solution**

**Data**

Initial velocity, u = 10ms^{-1}

Final velocity, v = 0

maximum height = displacement = 50m

Initial upward acceleration, a =?

When we apply the formula that says a = (v^{2} â€“ u^{2})/2d we will have

a = (0 â€“ 10^{2})/(2 x 50) = -100/100 = -1 ms^{-2}

Hence, since our acceleration is negative, we can now say that we are dealing with retardation or deceleration.

**Therefore, the retardation is -1ms ^{-2}**

**Note:** Retardation is the negative of acceleration, thus it is written in negative form.

### Problem 8

A car is traveling at a velocity of 8ms^{-1} and experiences an acceleration of 5ms^{-2}. How far does it travel in 4 seconds?

**Solution**

**Data:**

Initial velocity, u = 8ms^{-1}

acceleration, a = 5ms^{-2}

Distance, s =?

time, t = 4s

We can apply the formula that says

s = ut + (1/2)at^{2}

Thus

s = 8 x 4 + (1/2) x 5 x 4^{2}

s = 32 + (1/2) x 80 = 32 + 40 = 72m

**Therefore, the distance covered by the car in 4s is 72 meters. **

### Problem 9

A body is traveling at a velocity of 10m/s and experiences a deceleration of 5ms^{-2}. How long does it take the body to come to a complete stop?

**Solution**

**Data**

Initial velocity, u = 10m/s

Final velocity, v = 0

acceleration , a = retardation = -5ms^{-2}

time, t = ?

We already know that acceleration, a = change in velocity/time

This implies that

Time, t = change in velocity/acceleration

Thus

t = (v â€“ u)/t = (0-10)/-5 = -10/-5 = 2s

**Therefore, the time it takes the car to stop is 2 seconds**.

### Problem 10

A body is traveling at a velocity of 20 m/s and has a mass of 10 kg. How much force is required to change its velocity by 10 m/s in 5 seconds?

**Solution**

**Data**

Change in velocity, v =10 m/s

mass of the object, m = 10kg

time, t = 5s

We can apply newtonâ€™s second law of motion which says f = ma

and since a = change in velocity/time

we will have an acceleration equal to

a = 10/5 = 2ms^{-2}

Therefore, to find the force, we can now say

f = ma = 10 x 2 = 20N

**Therefore, the force that can help us to change the velocity by 10 m/s in 5 seconds is 20-Newton.**

Drop a comment if there is anything you donâ€™t understand about speed, velocity, and acceleration Problems.

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