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Understanding Motion through Experience - Frame of Reference-advanced

Grade 9CBSE

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

🔑Concepts

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A Frame of Reference is a coordinate system (like x,y,zx, y, z axes) combined with a clock, used by an observer to measure the position, velocity, and acceleration of an object.

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Motion is relative, not absolute. An object may appear to be in motion to one observer and at rest to another, depending on their respective frames of reference.

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An Inertial Frame of Reference is a frame that is either at rest or moving with a constant velocity. Newton's Laws of Motion are valid in these frames without any modification.

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A Non-Inertial Frame of Reference is a frame that is accelerating (either changing speed or changing direction). In such frames, Newton's Laws do not hold true in their standard form unless a pseudo force is considered.

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A Pseudo Force (or Fictitious Force) is an apparent force that acts on all masses whose motion is described using a non-inertial frame of reference, such as the centrifugal force experienced in a rotating frame.

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The Relative Velocity of object AA with respect to object BB is the velocity with which AA appears to move when BB is considered to be at rest.

📐Formulae

V⃗AB=V⃗A−V⃗B\vec{V}_{AB} = \vec{V}_A - \vec{V}_B

V⃗BA=V⃗B−V⃗A\vec{V}_{BA} = \vec{V}_B - \vec{V}_A

V⃗AB=−V⃗BA\vec{V}_{AB} = -\vec{V}_{BA}

Fpseudo=−m×aframeF_{pseudo} = -m \times a_{frame}

vrel=v1+v2 (Objects moving in opposite directions)v_{rel} = v_1 + v_2 \text{ (Objects moving in opposite directions)}

vrel=∣v1−v2∣ (Objects moving in the same direction)v_{rel} = |v_1 - v_2| \text{ (Objects moving in the same direction)}

💡Examples

Problem 1:

Two trains AA and BB are moving on parallel tracks. Train AA moves North at 60 km/h60\text{ km/h} and Train BB moves North at 45 km/h45\text{ km/h}. Calculate the velocity of Train AA relative to an observer in Train BB.

Solution:

Let North be the positive direction. Velocity of AA, VA=+60 km/hV_A = +60\text{ km/h} Velocity of BB, VB=+45 km/hV_B = +45\text{ km/h} Relative velocity VAB=VA−VBV_{AB} = V_A - V_B VAB=60−45=15 km/hV_{AB} = 60 - 45 = 15\text{ km/h}

Explanation:

Since both are moving in the same direction, the relative speed is the difference between their individual speeds. To a passenger in train BB, train AA appears to move North at 15 km/h15\text{ km/h}.

Problem 2:

A car XX is traveling East at 80 km/h80\text{ km/h} and another car YY is traveling West at 70 km/h70\text{ km/h}. What is the velocity of car XX with respect to car YY?

Solution:

Let East be positive (++) and West be negative (−-). VX=+80 km/hV_X = +80\text{ km/h} VY=−70 km/hV_Y = -70\text{ km/h} Relative velocity VXY=VX−VYV_{XY} = V_X - V_Y VXY=80−(−70)V_{XY} = 80 - (-70) VXY=80+70=150 km/hV_{XY} = 80 + 70 = 150\text{ km/h}

Explanation:

When objects move in opposite directions, their relative speeds add up. Car XX appears to be moving very fast (150 km/h150\text{ km/h}) to an observer in car YY.

Problem 3:

A person is standing in a bus that suddenly accelerates forward at 2 m/s22\text{ m/s}^2. If the mass of the person is 60 kg60\text{ kg}, calculate the magnitude of the pseudo force experienced by the person in the frame of the bus.

Solution:

Mass of the person, m=60 kgm = 60\text{ kg} Acceleration of the frame (bus), aframe=2 m/s2a_{frame} = 2\text{ m/s}^2 Fpseudo=m×aframeF_{pseudo} = m \times a_{frame} Fpseudo=60×2=120 NF_{pseudo} = 60 \times 2 = 120\text{ N}

Explanation:

In the non-inertial frame of the accelerating bus, the person experiences a pseudo force of 120 N120\text{ N} directed backward (opposite to the bus's acceleration).

Frame of Reference-advanced Class 9 Notes & Examples