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Motion in a Straight Line - Instantaneous Velocity and Speed

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Instantaneous velocity is defined as the velocity of an object at a particular instant of time tt. It is the limiting value of the average velocity as the time interval Δt\Delta t becomes infinitesimally small.

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Mathematically, instantaneous velocity vv is the first derivative of the position xx with respect to time tt, expressed as v=dxdtv = \frac{dx}{dt}.

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On a position-time (x−tx-t) graph, the instantaneous velocity at any point is equal to the slope of the tangent to the curve at that specific point.

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Instantaneous speed is the magnitude of the instantaneous velocity. While average speed and the magnitude of average velocity can differ, instantaneous speed is always equal to the magnitude of instantaneous velocity at that moment.

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The SI unit for both instantaneous velocity and speed is m/s\text{m/s} (or m s−1\text{m s}^{-1}).

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If the velocity of a particle is constant, its instantaneous velocity at any point is equal to its average velocity.

📐Formulae

v=lim⁡Δt→0ΔxΔtv = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}

v=dxdtv = \frac{dx}{dt}

Instantaneous Speed=∣v∣=∣dxdt∣\text{Instantaneous Speed} = |v| = \left| \frac{dx}{dt} \right|

Slope of x−t graph=tan⁡θ=dxdt\text{Slope of } x-t \text{ graph} = \tan \theta = \frac{dx}{dt}

💡Examples

Problem 1:

The position of an object moving along the xx-axis is given by x(t)=5t2+3t+4x(t) = 5t^2 + 3t + 4 (where xx is in meters and tt is in seconds). Find the instantaneous velocity of the object at t=3t = 3 s.

Solution:

Given the position function: x(t)=5t2+3t+4x(t) = 5t^2 + 3t + 4. To find the instantaneous velocity, we differentiate xx with respect to tt: v=dxdt=ddt(5t2+3t+4)v = \frac{dx}{dt} = \frac{d}{dt}(5t^2 + 3t + 4) Using power rule: v=10t+3v = 10t + 3 At t=3t = 3 s: v=10(3)+3=30+3=33 m/sv = 10(3) + 3 = 30 + 3 = 33 \text{ m/s}.

Explanation:

The instantaneous velocity is calculated by taking the derivative of the displacement-time equation and substituting the given value of time.

Problem 2:

A particle moves such that its displacement is x=4t3x = 4t^3. Calculate the instantaneous speed at t=2t = 2 s.

Solution:

The velocity is the derivative of displacement: v=dxdt=ddt(4t3)=12t2v = \frac{dx}{dt} = \frac{d}{dt}(4t^3) = 12t^2 At t=2t = 2 s: v=12(2)2=12×4=48 m/sv = 12(2)^2 = 12 \times 4 = 48 \text{ m/s} Since speed is the magnitude of velocity, the instantaneous speed is 48 m/s48 \text{ m/s}.

Explanation:

Instantaneous speed is the absolute value of the derivative of the position function. Since the result was positive, the magnitude remains the same.

Instantaneous Velocity and Speed - Revision Notes & Key Formulas | CBSE Class 11 Physics