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Physics - Motion in One Dimension (Scalar/Vector, Speed, Velocity, Acceleration)

Grade 9ICSE

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

🔑Concepts

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A Scalar quantity has only magnitude (e.g., massmass, distancedistance, speedspeed), whereas a Vector quantity has both magnitude and direction (e.g., displacementdisplacement, velocityvelocity, accelerationacceleration).

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Distance is the actual path length traveled by a body, while Displacement is the shortest straight-line distance between the initial and final positions in a specific direction.

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Speed is the rate of change of distance (Speed=DistanceTimeSpeed = \frac{Distance}{Time}), and it is always positive or zero.

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Velocity is the rate of change of displacement (Velocity=DisplacementTimeVelocity = \frac{Displacement}{Time}). It can be positive, negative, or zero.

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Acceleration (aa) is the rate of change of velocity with time. If the velocity increases, aa is positive; if the velocity decreases, it is called retardation or deceleration (negative aa).

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Uniform motion occurs when a body covers equal distances in equal intervals of time. Uniformly accelerated motion occurs when velocity changes by equal amounts in equal intervals of time.

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The slope of a Distance-Time graph represents SpeedSpeed, while the slope of a Displacement-Time graph represents VelocityVelocity.

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The slope of a Velocity-Time graph represents AccelerationAcceleration, and the area under a Velocity-Time graph represents DisplacementDisplacement.

📐Formulae

v=stv = \frac{s}{t}

Average Speed=Total DistanceTotal TimeAverage \ Speed = \frac{Total \ Distance}{Total \ Time}

a=v−uta = \frac{v - u}{t}

v=u+atv = u + at

s=ut+12at2s = ut + \frac{1}{2}at^2

v2=u2+2asv^2 = u^2 + 2as

sn=u+a2(2n−1)s_n = u + \frac{a}{2}(2n - 1)

💡Examples

Problem 1:

A car starts from rest and acquires a velocity of 54 km/h54 \text{ km/h} in 2 seconds2 \text{ seconds}. Find the acceleration of the car.

Solution:

Given: Initial velocity u=0 m/su = 0 \text{ m/s} (starts from rest). Final velocity v=54 km/hv = 54 \text{ km/h}. Converting to m/s\text{m/s}: v=54×518=15 m/sv = 54 \times \frac{5}{18} = 15 \text{ m/s}. Time t=2 st = 2 \text{ s}. Using the formula a=v−uta = \frac{v - u}{t}, we get a=15−02=7.5 m/s2a = \frac{15 - 0}{2} = 7.5 \text{ m/s}^2.

Explanation:

First, ensure all units are in the SI system (m/s\text{m/s}). Then apply the definition of acceleration as the change in velocity over time.

Problem 2:

A body moves with a constant acceleration of 2 m/s22 \text{ m/s}^2. If it starts from rest, calculate the distance covered by it in 5 seconds5 \text{ seconds}.

Solution:

Given: u=0 m/su = 0 \text{ m/s}, a=2 m/s2a = 2 \text{ m/s}^2, t=5 st = 5 \text{ s}. Using the second equation of motion: s=ut+12at2s = ut + \frac{1}{2}at^2. Substituting the values: s=(0)(5)+12(2)(5)2=0+25=25 ms = (0)(5) + \frac{1}{2}(2)(5)^2 = 0 + 25 = 25 \text{ m}.

Explanation:

Since the acceleration is uniform and we need to find the distance given time and initial velocity, the second equation of motion is the most appropriate.