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An athlete completes one round of circular track of diameter 50 m in 10 second. What will be the distance cover and the displacement at the end of 45 second?

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  An athlete completes one round of circular track of diameter 50 m in 10 second. What will be the distance cover and the displacement at the end of 45 second? Solution: Given: Diameter of circular track = 50 m So radius ( r ) = 25 m Time for 1 round = 10 s Total time = 45 s 1) Distance Covered First, find circumference of the track: Circumference = 2 π r = 2 π ( 25 ) = 50 π  m \text{Circumference} = 2\pi r = 2\pi (25) = 50\pi \text{ m} 50 π ≈ 157  m So, in 10 s → distance = 157 m In 45 s: Distance = 157 10 × 45  = 157 × 4.5 = 157 × 4.5                                  = 706.5  m (approx) Distance covered ≈ 706.5 m 2) Displacement After 45 Seconds In 40 seconds → athlete completes 4 full rounds (comes back to starting point). Remaining time = 5 seconds In 5 seconds → athlete completes half round. Half round means he reaches the diametrically opposite point . Displac...
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A body moves with an initial velocity of 2 m/s and uniform acceleration of 3 m/s2. Calculate the velocity when it has traveled a distance of 77 m.

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  A body moves with an initial velocity of 2 m/s and uniform acceleration of 3 m/s 2 . Calculate the velocity when it has traveled a distance of 77 m. Solution: Given: Initial velocity ( u ) = 2 m/s Acceleration ( a ) = 3 m/s² Displacement ( s ) = 77 m Since time is not given, we use the third equation of motion: v 2 = u 2 + 2 a s Step 1: Substitute values v 2 = ( 2 ) 2 + 2 ( 3 ) ( 77 ) v^2 = (2)^2 + 2(3)(77) v 2 = 4 + 6 × 77 v^2 = 4 + 6 × 77 v 2 = 4 + 462 v^2 = 4 + 462 v 2 = 466 v^2 = 466 Step 2: Take square root v = 466 v = \sqrt{466} ​ v ≈ 21.56  m/s
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A body starts from rest and moves with a uniform acceleration of 2 m/s2. What will be its velocity and displacement at the end of 10 seconds?

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  A body starts from rest and moves with a uniform acceleration of 2 m/s 2 . What will be its velocity and displacement at the end of 10 seconds? Given: Initial velocity ( u ) = 0 m/s (starts from rest) Acceleration ( a ) = 2 m/s² Time ( t ) = 10 s Velocity after 10 seconds Using the formula: v = u + a t v = u + at v = 0 + ( 2 × 10 ) v = 0 + (2 × 10) v = 20  m/s v = 20 \text{ m/s} Displacement after 10 seconds Using the formula: s = u t + 1 2 a t 2 s = ut + \frac{1}{2}at^2 s = 0 + 1 2 × 2 × ( 10 ) 2 s = 0 + \frac{1}{2} × 2 × (10)^2 s = 1 × 100 s = 1 × 100 s = 100  meters s = 100 \text{ meters}
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Inequality Notes class 8

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  Inequality Notes class 8 What is Inequality? Inequality refers to a mathematical statement that shows the relationship between two expressions that are not equal. In inequalities, the two expressions are connected using inequality symbols. Inequality Symbols: > (Greater Than) : Shows that the value on the left is larger than the value on the right. Example: 5 > 3 5 > 3 < (Less Than) : Shows that the value on the left is smaller than the value on the right. Example: 3 < 5 3 < 5 ≥ (Greater Than or Equal To) : Indicates that the value on the left is either greater than or equal to the value on the right. Example: 5 ≥ 3 5 \geq 3 ≤ (Less Than or Equal To) : Indicates that the value on the left is either less than or equal to the value on the right. Example: 3 ≤ 5 3 \leq 5 ≠ (Not Equal To) : Indicates that two values are not equal. Example: 5 ≠ 3 5 \neq 3 Solving Inequalities: Just like solving equations, inequalities can be solved to find th...
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An object dropped from a leaf false with a constant acceleration of 10 m/s2. Find its Speed 2 second after it was dropped.

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  An object dropped from a leaf false with a constant acceleration of 10 m/s 2 . Find its Speed 2 second after it was dropped.                            Solution: Given: Initial speed, u = 0 u = 0 (object is dropped) Acceleration due to gravity, a = 10   m/s 2 a = 10 \, \text{m/s}^2 Time, t = 2 \, \text{s} t  = 2 s Formula used: v = u + a t Calculation: v = 0 + ( 10 × 2 ) = 20   m/s Answer: The speed of the object after 2 seconds is 20 m/s (downwards) .
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A train travels 20 km at a uniform speed of 60 km/h and the next 20 km at a uniform speed of 80 km/hr. calculate its average speed.

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                                A train travels 20 km at a uniform speed of 60 km/h and the next 20 km at a uniform speed of 80 km/hr. calculate its average speed. Given: First distance d 1 = 20   km d_1 = 20 \, \text{km} , speed v 1 = 60   km/h v_1 = 60 \, \text{km/h} Second distance d 2 = 20   km d_2 = 20 \, \text{km} , speed v 2 = 80   km/h v_2 = 80 \, \text{km/h} Step 1: Time taken for each part t 1 = 20 60 = 1 3  h t_1 = \frac{20}{60} = \frac{1}{3} \text{ h} t 2 = 20 80 = 1 4  h t_2 = \frac{20}{80} = \frac{1}{4} \text{ h} Step 2: Total distance and total time Total distance = 20 + 20 = 40  km \text{Total distance} = 20 + 20 = 40 \text{ km} Total time = 1 3 + 1 4 = 4 + 3 12 = 7 12  h \text{Total time} = \frac{1}{3} + \frac{1}{4} = \frac{4 + 3}{12} = \frac{7}{12} \text{ h} Step 3: Average speed Average speed = Total distance Total time = 40 7 / 12 =...
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A boy is running on a straight road. He runs 500 m towards north in 2 minutes 10 seconds and then turns back and runs 200 m in 1 minute. Calculate i) His average speed and magnitude of average velocity during first 2 minutes 10 seconds and ii) His average speed and magnitude of average velocity during the whole journey.

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    A boy is running on a straight road. He runs 500 m towards north in 2 minutes 10 seconds and then turns back and runs 200 m in 1 minute. Calculate             i) His average speed and magnitude of average velocity during first 2 minutes 10 seconds and             ii) His average speed and magnitude of average velocity during the whole journey. Solution: Given: Distance covered towards north  = 500 m Time taken = 2 minutes 10 seconds = 130 s Then he runs back 200 m in 1 minute = 60 s (i) First 2 minutes 10 seconds This is only the first part of the motion. Distance covered = 500 m Time taken = 130 s (a) Average Speed :- Average Speed = Total Distance / Total Time = 500 / 130 = 3.85 m/s (b) Magnitude of Average Velocity :-      Displacement = 500 m towards north       Average velocity = Displacement / Time = 500 / 130 = 3.85 m/s (ii) Whole journey Total distance covered = 5...
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