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Motion - Motion in a Straight Line

Grade 9CBSE

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

🔑Concepts

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Motion is a relative term; an object may be in motion relative to one observer and at rest relative to another. A reference point or origin is needed to describe the position of an object.

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Distance is the total path length traveled by an object. It is a scalar quantity and is always positive.

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Displacement is the shortest distance between the initial and final positions of an object. It is a vector quantity and can be zero, positive, or negative. Magnitude of displacement ≤\le Distance.

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Uniform Motion: An object covers equal distances in equal intervals of time. The distance-time graph is a straight line passing through the origin.

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Non-Uniform Motion: An object covers unequal distances in equal intervals of time. The distance-time graph is a curve.

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Speed is the rate of change of distance (v=stv = \frac{s}{t}), while Velocity is the rate of change of displacement in a specific direction.

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Average Speed = Total distance travelledTotal time taken\frac{\text{Total distance travelled}}{\text{Total time taken}}.

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Acceleration (aa) is the rate of change of velocity per unit time. Its SI unit is m/s2\text{m/s}^2. If the velocity of an object increases, acceleration is positive; if it decreases, it is negative (Retardation).

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In a Velocity-Time graph, the area under the curve represents the displacement magnitude.

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Uniform Circular Motion: When an object moves in a circular path with uniform speed, its velocity changes at every point due to the change in direction. The speed is given by v=2πrtv = \frac{2\pi r}{t}.

📐Formulae

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

Average Velocity=u+v2\text{Average Velocity} = \frac{u + v}{2} (for uniform acceleration)

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

v=2πrtv = \frac{2\pi r}{t}

💡Examples

Problem 1:

A bus starting from rest moves with a uniform acceleration of 0.1 m/s20.1 \text{ m/s}^2 for 22 minutes. Find (a) the speed acquired, and (b) the distance travelled.

Solution:

Given: u=0 m/su = 0 \text{ m/s} (starts from rest), a=0.1 m/s2a = 0.1 \text{ m/s}^2, t=2 min=2×60=120 st = 2 \text{ min} = 2 \times 60 = 120 \text{ s}. (a) Using first equation: v=u+atv = u + at v=0+(0.1×120)=12 m/sv = 0 + (0.1 \times 120) = 12 \text{ m/s}. (b) Using second equation: s=ut+12at2s = ut + \frac{1}{2}at^2 s=(0×120)+12×0.1×(120)2s = (0 \times 120) + \frac{1}{2} \times 0.1 \times (120)^2 s=0+0.05×14400=720 ms = 0 + 0.05 \times 14400 = 720 \text{ m}.

Explanation:

We convert time to SI units (seconds) first. Since the bus starts from rest, initial velocity uu is zero. We then apply the equations of motion to find the unknown final velocity vv and distance ss.

Problem 2:

A farmer moves along the boundary of a square field of side 10 m10 \text{ m} in 40 s40 \text{ s}. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds2 \text{ minutes } 20 \text{ seconds} from his initial position?

Solution:

Side of square =10 m= 10 \text{ m}. Perimeter =4×10=40 m= 4 \times 10 = 40 \text{ m}. Time for 1 round =40 s= 40 \text{ s}. Total time =2 min 20 s=140 s= 2 \text{ min } 20 \text{ s} = 140 \text{ s}. Number of rounds =14040=3.5= \frac{140}{40} = 3.5 rounds. After 3.53.5 rounds, the farmer is at the opposite corner of the square field. Displacement =Diagonal of the square=102+102=200=102 m= \text{Diagonal of the square} = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2} \text{ m}. Displacement ≈14.14 m\approx 14.14 \text{ m}.

Explanation:

Displacement is the straight-line distance between start and end. Since 3.53.5 rounds means 33 full rounds plus half a round, the farmer ends up at the diagonally opposite vertex. We use Pythagoras theorem to find this diagonal distance.

Motion in a Straight Line Class 9 Notes & Examples