Speed, velocity, acceleration

Welcome back! In the last section, we built our foundation by understanding frames of reference, distance, and displacement. Now, we'll build upon that to describe how fast and in what direction an object is moving. This leads us to the core concepts of kinematics: speed, velocity, and acceleration.

These three terms are often used interchangeably in everyday language, but in physics, they have precise and distinct meanings. Mastering this distinction is the first major step towards cracking numerical problems in kinematics. The table above is your first go-to reference.

Speed: The Measure of "How Fast"

Speed is a scalar quantity that tells us how fast an object is moving. It's simply the rate of change of distance. Since distance is a scalar and time is a scalar, their ratio (speed) is also a scalar. It has no direction associated with it. When your car's speedometer reads "60 km/h", it's indicating your speed.

The formula for speed is straightforward:

Speed = Distance / Time

The SI unit for speed is meters per second (m/s). A common practical unit is kilometers per hour (km/h).

{CALLOUT: type=tip | text=Exam-Ready Conversion: Remember this conversion by heart! To convert from km/h to m/s, multiply by 5/18. To convert from m/s to km/h, multiply by 18/5. <br> Example: 90 km/h = 90 × (5/18) m/s = 5 × 5 = 25 m/s.}

Types of Speed

We can classify speed into four main types, each crucial for solving different kinds of problems.

1. Uniform Speed: An object has uniform speed if it covers equal distances in equal intervals of time, no matter how small these intervals are. The motion of the tip of a clock's second hand is an example of uniform speed.

2. Non-uniform or Variable Speed: If an object covers unequal distances in equal intervals of time, its speed is non-uniform. A car moving in city traffic is a perfect example; it speeds up, slows down, and stops.

3. Average Speed: This is the most frequently tested concept in NEET. For an object moving with variable speed, the average speed is the ratio of the total distance travelled to the total time taken. It gives a single value to represent the entire journey's rate of motion.

{FORMULA: expr=v_avg = (Total Distance Travelled) / (Total Time Taken) | symbols=v_avg: Average speed}

4. Instantaneous Speed: This is the speed of an object at a particular instant of time. Think of it as the average speed over an infinitesimally small time interval. Your car's speedometer shows your instantaneous speed. Mathematically, if Δs is the distance travelled in a very small time interval Δt, then: Instantaneous Speed, v = lim (Δs/Δt) as Δt → 0. This is simply the magnitude of the instantaneous velocity.

Numerical Examples on Average Speed

Average speed problems are a favourite of examiners. Let's tackle the most common patterns.

Example 1: Journey in two equal time intervals

A car travels on a straight road. For the first half of the time of the journey, it travels with a speed v₁, and for the second half of the time, it travels with speed v₂. Find the average speed of the car.

Given:

• Speed in first half-time = v₁
• Speed in second half-time = v₂
• Time interval 1 (t₁) = Time interval 2 (t₂) = T/2, where T is the total time.

To Find:

• Average speed (v_avg)

Approach: We'll use the fundamental formula for average speed: v_avg = Total Distance / Total Time. We need to calculate the total distance by finding the distance covered in each time interval.

Working:

1. Let the total time of the journey be T.
2. The first time interval is t₁ = T/2. The distance covered is d₁ = v₁ × t₁ = v₁ × (T/2).
3. The second time interval is t₂ = T/2. The distance covered is d₂ = v₂ × t₂ = v₂ × (T/2).
4. Total distance D = d₁ + d₂ = (v₁T/2) + (v₂T/2) = (v₁ + v₂) × (T/2).
5. Total time taken is T.
6. Average speed v_avg = D / T.

v_avg = [(v₁ + v₂) × (T/2)] / T

v_avg = (v₁ + v₂) / 2

Final Answer: The average speed is the arithmetic mean of the two speeds: (v₁ + v₂)/2.

Example 2: Journey in two equal distance intervals

This is a classic NEET question pattern. A particle covers half of its total distance with speed v₁ and the rest half distance with speed v₂. Find its average speed during the complete journey.

{{SOLVE: {"problem":"A particle covers half of its total distance with speed v₁ and the rest half distance with speed v₂. Find its average speed.","type":"derivation","subject":"physics","intro":"Yeh ek bahut important aur common question hai. Let's solve it on the whiteboard step-by-step.","outro":"So, the average speed is the harmonic mean. Ab isse direct formula ki tarah yaad rakho!","steps":[{"explanation":"First, let's define our variables. Let the total distance be 2d. This makes the two halves simply 'd' each.","write":"Let Total Distance = 2d. So, d₁ = d and d₂ = d.","tough":false},{"explanation":"Now, we need the total time. Time is distance divided by speed. Let's find the time taken for each half of the journey, t₁ and t₂.","write":"Time for first half, t₁ = d₁/v₁ = d/v₁","tough":false},{"explanation":"Similarly, let's find the time for the second half of the journey.","write":"Time for second half, t₂ = d₂/v₂ = d/v₂","tough":false},{"explanation":"The total time for the journey is simply the sum of the individual times, T = t₁ + t₂.","write":"Total Time, T = t₁ + t₂ = (d/v₁) + (d/v₂)","tough":true,"alt_explanation":"We just add the two time expressions we found in the previous steps."},{"explanation":"To add these fractions, we need a common denominator, which is v₁v₂.","write":"T = d(1/v₁ + 1/v₂) = d( (v₂ + v₁) / (v₁v₂) )","tough":true,"alt_explanation":"We're taking 'd' common and then finding the LCM of the denominators to combine the fractions."},{"explanation":"Now we have the total distance (2d) and the total time. Let's plug them into the average speed formula: v_avg = Total Distance / Total Time.","write":"v_avg = (2d) / [ d(v₁ + v₂) / (v₁v₂) ]","tough":false},{"explanation":"The 'd' in the numerator and denominator will cancel out. Now we just need to simplify the expression.","write":"v_avg = 2 / [ (v₁ + v₂) / (v₁v₂) ]","tough":true,"alt_explanation":"Notice how the 'd' cancels. The result is independent of the actual distance. Now, we flip the fraction in the denominator and multiply."},{"explanation":"Flipping the denominator and multiplying gives us the final, clean formula for the average speed.","write":"v_avg = 2v₁v₂ / (v₁ + v₂)","tough":false}]}}}

The result v_avg = 2v₁v₂ / (v₁ + v₂) is the Harmonic Mean of the two speeds. This is a very important result to remember directly for MCQs.

Velocity: Speed with Direction

While speed tells us "how fast", velocity tells us how fast and in which direction. It is a vector quantity defined as the rate of change of displacement.

{{VISUAL: diagram: A car traveling from point A to point B along a winding road. The total length of the road is labeled as 'Distance (path length)'. A straight arrow pointing from A to B is labeled 'Displacement (shortest path, has direction)'.}}

The formula for average velocity is:

Average Velocity (v⃗_avg) = Total Displacement (Δr⃗) / Total Time Taken (Δt)

The arrow (⃗) above the symbols indicates that they are vector quantities. The SI unit for velocity is the same as speed: m/s.

Types of Velocity

1. Uniform Velocity: An object has uniform velocity if it covers equal displacements in equal intervals of time. This implies that both its speed and its direction of motion are constant. Motion with uniform velocity is always motion in a straight line.

2. Non-uniform or Variable Velocity: If an object's speed, or its direction of motion, or both, change with time, it is said to be moving with variable velocity. A car turning a corner, even at a constant speed, has a changing velocity because its direction is changing. An object in uniform circular motion has constant speed but variable velocity.

3. Average Velocity: This is the ratio of the total displacement to the total time interval. It only cares about the starting and ending points, not the path taken in between.

4. Instantaneous Velocity: This is the velocity of an object at a specific instant in time. It is defined as the time derivative of the position vector. v⃗ = lim (Δr⃗/Δt) as Δt → 0, which in calculus notation is:

   v⃗ = dr⃗/dt
   

   The direction of the instantaneous velocity is always tangent to the path of the particle at that point. The magnitude of the instantaneous velocity is the instantaneous speed. |v⃗| = v.

{KEY: type=exam | title=Average Speed vs. Average Velocity: The Ultimate Test Case | text=Consider an athlete running one full circle on a circular track of radius 'r' in time 't'.

• Total Distance = Circumference = 2πr
• Total Displacement = 0 (since the athlete returns to the starting point)
• Average Speed = (2πr) / t (This is a non-zero value!)
• Average Velocity = 0 / t = 0 (This is zero!) This single example clarifies the difference perfectly. Average speed can be non-zero even when average velocity is zero.}

Example 3: Comparing Average Speed and Velocity

A man walks 30 m North, then 40 m East, in a total time of 50 seconds. Calculate his (a) average speed and (b) average velocity.

Given:

• Distance North (d₁) = 30 m
• Distance East (d₂) = 40 m
• Total time (t) = 50 s

To Find:

• (a) Average speed
• (b) Average velocity

Approach: For average speed, we need the total path length (distance). For average velocity, we need the net displacement (the straight-line distance from start to finish).

{{VISUAL: diagram: A coordinate system with the origin as the starting point. An arrow points 30 units up along the y-axis (North). From the tip of that arrow, another arrow points 40 units right along the x-axis (East). A third, dashed arrow connects the origin to the final point, forming the hypotenuse of a right-angled triangle. This hypotenuse is labeled 'Displacement'.}}

Working: (a) Average Speed

1. The path is 30 m followed by 40 m.
2. Total Distance = d₁ + d₂ = 30 m + 40 m = 70 m.
3. Total Time = 50 s.
4. Average Speed = Total Distance / Total Time.

v_avg_speed = 70 m / 50 s = 1.4 m/s

(b) Average Velocity

1. The two parts of the journey are perpendicular. The displacement is the hypotenuse of a right-angled triangle with sides 30 m and 40 m.
2. Magnitude of Displacement = √(30² + 40²) = √(900 + 1600) = √(2500) = 50 m.
3. Average Velocity = Total Displacement / Total Time.

v_avg_velocity_magnitude = 50 m / 50 s = 1.0 m/s

4. The direction is from the starting point to the final point. The angle θ it makes with the East direction is tan(θ) = Opposite/Adjacent = 30/40 = 3/4. So, θ = tan⁻¹(3/4) North of East.

Final Answer:

• (a) The average speed is 1.4 m/s.
• (b) The average velocity is 1.0 m/s at an angle of tan⁻¹(3/4) North of East.

Acceleration: The Rate of Change of Velocity

Whenever the velocity of an object changes, we say it is accelerating. Acceleration is the rate of change of velocity with respect to time. Since velocity is a vector, acceleration is also a vector.

Its SI unit is meters per second squared (m/s²).

The formula for average acceleration is:

Average Acceleration (a⃗_avg) = Change in Velocity (Δv⃗) / Total Time Taken (Δt)

a⃗_avg = (v⃗_final - v⃗_initial) / (t_final - t_initial)

{CALLOUT: type=warning | text="Retardation" is just negative acceleration! In physics, we prefer the term acceleration for all changes in velocity. If an object is slowing down, its acceleration vector points in the opposite direction to its velocity vector. This is often called deceleration or retardation, but it's simply a negative acceleration relative to the chosen positive direction.}

Types of Acceleration

1. Uniform Acceleration: If the velocity of an object changes by equal amounts in equal intervals of time, the acceleration is uniform. A body falling freely under gravity (neglecting air resistance) is the classic example. The equations of motion (v = u + at, etc.) are only valid for uniform acceleration.

2. Non-uniform or Variable Acceleration: If the change in velocity is different in equal time intervals, the acceleration is variable. The motion of a car on a crowded street involves variable acceleration.

{{VISUAL: chart: Two Velocity-Time (v-t) graphs side-by-side. (a) Labeled 'Uniform Acceleration'. Shows a straight line with a positive, constant slope. The slope is labeled 'a = constant > 0'. (b) Labeled 'Non-uniform Acceleration'. Shows a curve that is getting steeper. The slope is labeled 'a is increasing'.}}

3. Instantaneous Acceleration: This is the acceleration of an object at a given instant. It is the time derivative of velocity, or the second time derivative of position. a⃗ = lim (Δv⃗/Δt) as Δt → 0, which in calculus notation is:

   a⃗ = dv⃗/dt = d²r⃗/dt²
   

Example 4: Calculating Instantaneous Acceleration

The position of a particle moving along the x-axis is given by x = 3t³ - 6t² + 4t + 5, where x is in meters and t is in seconds. Find the acceleration of the particle at t = 2 s.

Given:

• Position function: x(t) = 3t³ - 6t² + 4t + 5

To Find:

• Acceleration a at t = 2 s.

Approach: Acceleration is the second derivative of position with respect to time (a = d²x/dt²). We must differentiate the position function twice.

Working:

1. First, find the velocity function v(t) by differentiating x(t) with respect to t. v(t) = dx/dt = d/dt (3t³ - 6t² + 4t + 5)

   v(t) = 9t² - 12t + 4
   

2. Now, find the acceleration function a(t) by differentiating v(t) with respect to t. a(t) = dv/dt = d/dt (9t² - 12t + 4)

   a(t) = 18t - 12
   

3. Finally, substitute t = 2 s into the acceleration function a(t). a(2) = 18(2) - 12

   a(2) = 36 - 12 = 24 m/s²
   

Final Answer: The acceleration of the particle at t = 2 s is 24 m/s².

Problem-Solving Zone

This section is designed to sharpen your skills for the NEET exam.

Common Numerical Traps

NEET-Style MCQ Bank

1. A car covers the first half of the distance between two places at 40 km/h and the other half at 60 km/h. The average speed of the car is: (a) 40 km/h (b) 48 km/h (c) 50 km/h (d) 60 km/h

   💡 Answer: (b) Solution: This is a case of equal distances. We use the harmonic mean formula: v_avg = 2v₁v₂ / (v₁ + v₂). v_avg = (2 × 40 × 60) / (40 + 60) = 4800 / 100 = 48 km/h. Option (c) 50 km/h is the arithmetic mean, (40+60)/2, which is a common mistake.

2. A particle's position is described by the equation x = (t² - 5t + 6) meters. The time when the particle's velocity is zero is: (a) 2 s (b) 3 s (c) 0 s (d) 2.5 s

   💡 Answer: (d) Solution: First, find the velocity function v = dx/dt. v = d/dt (t² - 5t + 6) = 2t - 5. Now, set the velocity to zero to find the time: 2t - 5 = 0. 2t = 5 → t = 2.5 s.

3. An object travels 10 km at a speed of 100 m/s and the next 10 km at a speed of 50 m/s. The average speed over the whole distance is approximately: (a) 75 m/s (b) 66.7 m/s (c) 55 m/s (d) 82.5 m/s

   💡 Answer: (b) Solution: This is another case of equal distances. We can use the harmonic mean formula v_avg = 2v₁v₂ / (v₁ + v₂). v_avg = (2 × 100 × 50) / (100 + 50) = 10000 / 150 = 1000 / 15 = 200 / 3 = 66.67 m/s. Option (a) is the incorrect arithmetic mean.

4. If a car at rest accelerates uniformly to a speed of 144 km/h in 20 s, it covers a distance of: (a) 20 m (b) 400 m (c) 1440 m (d) 2880 m

   💡 Answer: (b) Solution: First, convert speed to SI units. v = 144 km/h = 144 × (5/18) = 8 × 5 = 40 m/s. Initial velocity u = 0. Time t = 20 s. We can use the equation of motion s = (u+v)/2 × t. s = (0 + 40)/2 × 20 = 20 × 20 = 400 m. Alternatively, find a = (v-u)/t = (40-0)/20 = 2 m/s². Then use s = ut + ½at² = 0 + ½(2)(20)² = 400 m.

Practice Set

Test your understanding with these problems.

1. A body travels from A to B at 40 m/s and from B to A at 60 m/s. Calculate the average speed and average velocity.

   💡 Answer: Average Speed = 48 m/s, Average Velocity = 0.

2. The displacement of a particle is given by y = a + bt + ct² - dt⁴. Find the initial velocity and initial acceleration.

   💡 Answer: Initial Velocity = b, Initial Acceleration = 2c.

3. A person travels along a straight road for the first half time with a velocity v₁ and the next half time with a velocity v₂. What is the mean velocity of the person?

   💡 Answer: v_mean = (v₁ + v₂)/2.

4. A cheetah can accelerate from rest to 72 km/h in 2 seconds. What is its average acceleration?

   💡 Answer: 10 m/s².

5. The velocity of a particle is given by v = 2t² - 4t + 3. Find its acceleration at t = 3 s.

   💡 Answer: 8 m/s².

Quick Recap Cheatsheet

{FLASHCARD: q=An object is in uniform circular motion. Is it accelerating? | a=Yes. Although its speed is constant, its direction of velocity is continuously changing. A change in velocity (even just direction) means there must be an acceleration (centripetal acceleration).}