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Understanding Motion through Experience - What is Motion?-advanced

Grade 9CBSE

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

🔑Concepts

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Motion is defined as a change in the position of an object over time with respect to a fixed Reference Point (also called the Origin). Motion is relative; an object may be in motion relative to one observer but at rest relative to another.

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Distance is a scalar quantity representing the total path length covered by an object, while Displacement is a vector quantity representing the shortest straight-line distance between the initial and final positions.

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Uniform Motion occurs when an object covers equal distances in equal intervals of time, regardless of how small these intervals may be. Non-Uniform Motion occurs when the distance covered varies over equal time intervals.

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Speed is the rate of change of distance (scalar), while Velocity is the rate of change of displacement (vector). Velocity includes both magnitude (speed) and direction.

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Acceleration (aa) is the measure of the change in velocity of an object per unit time. If velocity increases, acceleration is positive; if velocity decreases, it is called deceleration or retardation (negative acceleration).

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On a Distance-Time Graph, the slope represents the speed of the object. On a Velocity-Time Graph, the slope represents acceleration, and the area under the curve represents the displacement.

📐Formulae

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

v⃗avg=Total DisplacementTotal Time\vec{v}_{avg} = \frac{\text{Total Displacement}}{\text{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

vavg=u+v2 (for uniform acceleration)v_{avg} = \frac{u + v}{2} \text{ (for uniform acceleration)}

💡Examples

Problem 1:

An athlete completes one round of a circular track of diameter 200 m200\text{ m} in 40 s40\text{ s}. What will be the distance covered and the displacement at the end of 2 minutes 20 seconds2\text{ minutes } 20\text{ seconds}?

Solution:

Given: Diameter d=200 md = 200\text{ m}, so Radius r=100 mr = 100\text{ m}. Time for 1 round =40 s= 40\text{ s}. Total time =2×60+20=140 s= 2 \times 60 + 20 = 140\text{ s}. Number of rounds =14040=3.5 rounds= \frac{140}{40} = 3.5\text{ rounds}. Distance =Number of rounds×2πr=3.5×2×227×100=2200 m= \text{Number of rounds} \times 2\pi r = 3.5 \times 2 \times \frac{22}{7} \times 100 = 2200\text{ m}. After 3.53.5 rounds, the athlete is at the diametrically opposite point from the start. Displacement =Diameter=200 m= \text{Diameter} = 200\text{ m}.

Explanation:

Distance is the actual path (circumference ×\times rounds), while displacement depends only on the final position relative to the starting position.

Problem 2:

A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h40\text{ km/h} in 10 minutes10\text{ minutes}. Find its acceleration in m/s2m/s^2.

Solution:

Initial velocity u=0 m/su = 0\text{ m/s} (starts from rest). Final velocity v=40 km/h=40×518=1009 m/s≈11.11 m/sv = 40\text{ km/h} = 40 \times \frac{5}{18} = \frac{100}{9}\text{ m/s} \approx 11.11\text{ m/s}. Time t=10 min=10×60=600 st = 10\text{ min} = 10 \times 60 = 600\text{ s}. Using a=v−uta = \frac{v - u}{t}: a=11.11−0600a = \frac{11.11 - 0}{600} a≈0.0185 m/s2a \approx 0.0185\text{ m/s}^2

Explanation:

Acceleration is the change in velocity over time. We must convert units to the SI system (m/sm/s and secondsseconds) before calculating.

Problem 3:

A bus starting from rest moves with a uniform acceleration of 0.1 m/s20.1\text{ m/s}^2 for 2 minutes2\text{ minutes}. Find (a) the speed acquired, and (b) the distance travelled.

Solution:

Given: u=0u = 0, a=0.1 m/s2a = 0.1\text{ m/s}^2, t=120 st = 120\text{ s}. (a) To find vv: v=u+atv = u + at v=0+(0.1×120)=12 m/sv = 0 + (0.1 \times 120) = 12\text{ m/s} (b) To find ss: 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=0.05×14400=720 ms = 0.05 \times 14400 = 720\text{ m}

Explanation:

The equations of motion are used here. Part (a) uses the first equation to find final velocity, and part (b) uses the second equation to find the distance covered during acceleration.