Relative Speed: Explanation, Formula and Examples

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Relative speed is the speed of a moving body in relation to another. The word “Relative” can be simplified as “in comparison with”. When two bodies are moving in the same direction, their difference is used to calculate their relative speed. When two bodies are moving in opposite directions, however, the relative speed is derived by summing their speeds.

Table of Content

What is Speed?
What is Relative Speed?
Formula for Relative Speed
Determination of Relative Speed
Difficulties of Calculating Relative Speed
Things to Remember
Sample Questions

Key Terms- Speed, Motion, Velocity, Relative motion, Directions, Vector Quality, Formulas

What is Speed?

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The distance travelled per unit of time is known as speed. It is the rate at which a body moves, that is how fast someone is coming towards us or going away from us. Speed is a scalar quantity meaning, there is no direction to speed. On the other hand, velocity is a vector quantity.

What is Relative Speed?

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When two or more bodies travelling at different speeds are evaluated, the idea of relative speed is applied. To make things easier, one body can be made stationary (i.e. Speed = 0), and the other body's speed with regard to the stationary body can be calculated as the total of the speeds if the bodies are travelling in different directions, or as the difference, if they are going in the same direction. Relative Speed is the speed of a moving body in relation to a stationary body.

If two bodies are going in opposite directions, their relative speed is equal to the sum of their individual speeds.
If they are moving in the same direction, their relative speed is equal to the difference in their individual speeds.
The difference between relative speed and relative velocity is that the former is a scalar quantity, while the latter is a vector quantity.

Also Read:

Difference Between Speed and Velocity

Distance and Displacement

Derivation of Equation of Motion

Formula of Relative Speed

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The formula below shows how to compute the constant velocity of an item travelling in a straight line using the traditional method.

r = \(\frac{d}{t}\)

r → Rate or speed

d→ Distance travelled

t → Amount of time it takes to execute an activity

The above equation determines the average speed of an object over a period of time. At different moments over the time span, the item may be moving faster or slower. The entire distance travelled by an object divided by the total time taken is the average speed.

If two bodies are moving at different speeds in the same direction, then their relative speed can be determined by the difference between their speeds. It can be expressed by

V₁ - V₂

If two bodies are moving at different speeds in the opposite direction, then their relative speed can be determined by the summation of their speeds. It can be expressed by

V₁ + V₂

Where,

V₁ → Speed of the first body

V₂ → Speed of the second body

Also Read: Speed Time Graphs

Determination of Relative Speed

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Here is how we can determine the Relative Speed-

A and B are two bodies travelling towards each other at 20 kmph and 30 kmph, respectively. Make body A immobile and calculate B's speed in relation to A, thus B's relative speed Equals the total of their respective speeds = 20 + 30 = 50 kmph ( as they are travelling in opposite directions).

Assume that two bodies are going in the same direction but at different speeds.

Let the first body's speed be x km/hr and the second body's speed be y km/hr.

As a result, their relative speed is equal to (x – y) km/hr where [x > y].

Then,

The time it takes for two bodies to collide = distance travelled / relative speed

→ d km / (x – y) km/hr

We already know that relative speed refers to how fast one body moves in relation to another.

Assume that time is equal to t hours.

The distance reached in ‘t' hours is then equal to relative speed x time.

→ (x – y) km/hr * t hrs

Also Read:

Uniform Acceleration

Kinematics Equations

Average Acceleration Formula

Difficulties of Determining Relative Speed

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Here are some situations where calculating the Relative Speed can be tricky-

Meeting Time and Travel Time: If two moving body has started to moves towards each other at the same time maintaining the same speed, calculating their meeting point can be tricky. Though the distances covered by them should remain proportional to their speed.

Issues Related to the Ratio: When two moving bodies are following the same path, determining Relative Speeds may lead to some ration-based errors. For example- X body and Y body are moving in the same direction with the similar speed and the speed and distance is mentioned as X:Y. But it can also be Y:X as per the time to cover the same distance. But in this case the distance, speed and time must remain constant.

Traveling By Boat: In this type of calculation the speed of the river flow must be taken into consideration. But the speed of the river flowing can not be counted as a constant. For example, a boat can travel a certain distance but the situations like upstreams and downstreams will affect the calculation.

Also Read: Difference Between Kinetics and Kinematics

Things to Remember

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Traveling by Boat: In this type of calculation the speed of the river flow must be taken into consideration. But the speed of the river flowing can not be counted as a constant. For example, a boat can travel a certain distance but the situations like upstream and downstrems will affect the calculation.
- Relative speed is the speed of a moving body in relation to another.
- When two bodies are moving in the same direction, their difference is used to calculate their relative speed.
- It is given by V₁ - V₂. Where V₁ and V₂ are the speed of the bodies.
- When two bodies are moving in opposite directions, the relative speed is derived by summing their speeds.
- It is given by V₁ + V₂. Where V₁ and V₂ are the speed of the bodies.

Also Have a Look at the PYQs for Reference:

In vertical circular motion, the ratio of kinetic energy of a particle…. (MHT CET 2016)
The motion of a particle in straight line is…. (JKCET 2013)
A charged particle enters a uniform magnetic field with a certain speed …. (JKCET 2006)
From the top of a tower a body AA is projected vertically up…. (JKCET 2007)
Assuming earth to be an inertial frame…. (JKCET 2009)
A body is sliding on a smooth inclined plane requires…. (JIPMER 1999)
A body of mass 5kg is suspended by a spring balance on… (BITSAT 2006)
A body of mass M hits normally a rigid wall with velocity V … (BITSAT 2018)
A man of weight 80kg is standing in an elevator which is… (DUET 2003)
A light inextensible string that goes over a smooth fixed pulley … (BITSAT 2013)

Sample Questions

Ques. Pakad Singh, a police officer, spots the thief Bhagu Ram from a distance of 200 metres. They start running as soon as they see one other. What is the distance Bhagu Ram would have covered at 5 kmph before being captured by Pakad Singh at 7 kmph? (3 marks)

Ans. Two bodies are travelling at separate rates. Assume Bhagu Ram is stationary and calculate Pakad Singh's speed in relation to Bhagu Ram. So, using the notion of relative speed, it can be determined that Bhagu Ram is stationary and Pakad Singh is running at 7 – 5 = 2 kmph at a distance of 200 m. He must travel 200 metres or 0.2 kilometres to catch Bhagu Ram. As a result, the time taken is 0.2/2 = 0.1 hour or 6 minutes. As a result, Bhagu Ram travels 0.5 kilometres before being captured.

Ques. A bird is seated on train A, which is travelling at a speed of 40 kilometres per hour. It notices another train B approaching from the other direction on the same rail track at a distance of 200 metres and a speed of 60 kilometres per hour. It travels at a pace of 10 kilometres per hour and takes a seat on another train. It quickly returns to the first train, then to the second train, and so on. Before the two trains collide, it does so. What is the total distance covered by the bird? (3 marks)

Ans. Two trains are travelling in opposing directions at speeds of 40 and 60 kilometres per hour. Make train B stop, then calculate the speed of train A in relation to train B. As a result, train A's relative speed is 40 + 60 = 100 kmph.

It must go 200 metres before colliding with the stationary train B, which it will do in (0.2/100) hours. The average speed of the bird during this time was 10 km/h, implying that the total distance reached by the bird before the collision was 0.2 * 10/100 km = 20 metres.

Ques. Gita rows a boat upstream at 15 kmph and downstream at 20 kmph. Determine the speed with which Gita rows the boat in still water, as well as the stream's speed. (2 marks)

Ans. Given that upstream Speed = 15 kmph

Downstream Speed = 20 kmph

Speed of Gita in still water = x = ((a + b))/2= (20+15)/2 = 35/2

Speed of stream =y = ((a – b))/2= (20-15)/2 = 5/2.

Ques. The Narmada River travels at a speed of 5 kilometres per hour. In the river, a stationary body is placed. Calculate the time it took the floating body to reach a stone 10 kilometres downstream from where it is now. (2 marks)

Ans. Speed of body = Speed of river (as Speed of boy is 0) = 5 kmph

Speed=Distance/Time.

So, Time taken to reach 10 km = 10/5 = 2 hours.

Ques. In 2.7 hours, a guy can row 135 kilometres upstream. In 2.5 hours, he can row the same distance downstream. However, he reduces his downstream speed by 9%, while the current speed is cut by 20%. Discover the man's speed. (2 marks)
(a) 52
(b) 55
(c) 50
(d) none of these

Ans. For this question, let's take an alternative approach: x= The boat's speed and

y= The Stream's Speed

x-y= 135/2.7= 50 —–(1) & 10/11 x + 4/5 y = 135/2.5 = 54 ——– (2)

(A 9 percent reduction equals 100- 9 = 0.91x0.91, which can be expressed as 10/11 because 1/11= 0.0909). (Similarly, a 20% reduction => 0.8 y.0.8, which can be represented as 4/5 in fraction form.)

Option (c)The man's speed cannot be 50 since the stream's speed cannot be zero (when substituting in equation (1)).

Option (b) In the second equation, substitute x= 52 and x=55 to determine which value gives you an integral value for y. This is only true when x=55, where y=5 is the result. As a result, the option is the answer (b) The Reverse Gear method is applied here, which includes working from answer options. It's important to remember that the longer the question, the easier it is to remove the incorrect answers.

Ques. In two and a half minutes, a runner may complete a 750-metre race. Will he be able to beat a runner who clocks in at 17.95 kilometres per hour? (2 marks)

Ans. We're told that the first runner can run a 750-meter race in 2 minutes, 30 seconds, or 150 seconds.

=> The first runner's speed is 750/150 = 5 metres per second.

We calculate this speed by multiplying it by 18/5 to get kilometres per hour.

=> The first runner's speed is 18 km/hr.

We also know that the second runner's top speed is 17.95 km/hr.

As a result, the first runner has a chance to beat the second runner.

Ques. What is Relative Velocity? (2 marks)

Ans. The relative velocity of object A with respect to another object B is the time rate at which object A changes its position with respect to object B.

—> The relative velocity of two objects moving in the same direction is the difference of the speeds of the objects.

—> The relative velocity of two objects moving in the opposite direction is the sum of the speeds of the objects.

Ques. A jet airplane travelling at the speed of 500 km h^-1 ejects its products of combustion at the speed of 1500 km h?¹ relative to the jet plane. What is the speed of the latter with respect to an observer on the ground? (2 marks)

Ans. Velocity of jet airplane w.r.t observer on ground = 500 km/h.

If v_j and v₀ represent the velocities of jet and observer respectively, then v_j – v_o = 500 km h^-1

Similarly, if v_c represents the velocity of the combustion products w.r.t jet plane, then v_c – v_g = -1500 km/h

The negative sign indicates that the combustion products move in a direction opposite to that of jet.

Speed of combustion products w.r.t. observer

= v_c – u₀ = (v_c – v_j) + (v_j – v₀) = (-1500 + 500) km h^-1 = -1000 km h^-1

Also Read,

Path Length	Wien’s displacement Law	Newton’s Third law of motion
Vector and Scalar Quantities	Displacement Formula	Heights and Distances
Angular speed	Distance-Time Graph	Oscillations

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Relative Speed: Explanation, Formula and Examples

What is Speed?

What is Relative Speed?

Formula of Relative Speed

Determination of Relative Speed

Difficulties of Determining Relative Speed

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.
Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

2.
If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

3.
Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

5.
Write any two features of nuclear forces.

6.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

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Relative Speed: Explanation, Formula and Examples

What is Speed?

What is Relative Speed?

Formula of Relative Speed

Determination of Relative Speed

Difficulties of Determining Relative Speed

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

2. If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

3.Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

5. Write any two features of nuclear forces.

6.Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

2.
If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

3.
Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

5.
Write any two features of nuclear forces.

6.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.