Motion Class 9 Notes: Race - PDF

Here is the transcription of the document in markdown format: # RACE CLASS 9TH ONE SHOT Physics MOTION By - Akshay Tyagi Sir ### Topics to be covered 1. One Shot Revision 2. Numerical ### Motion and rest * Motion and rest are relative * REST: An object that does not change its positions wi...

Here is the transcription of the document in markdown format: # RACE CLASS 9TH ONE SHOT Physics MOTION By - Akshay Tyagi Sir ### Topics to be covered 1. One Shot Revision 2. Numerical ### Motion and rest * Motion and rest are relative * REST: An object that does not change its positions with time is said to be at rest with respect to Observer * MOTION: An object that changes its positions with time is said to be in motion wr.t Observer ### Physical Quantities * Physical Quantities: that can be measured * Scalar Quantities: Physical quantities which can be completely described by their magnitude alone are called 'Scalar Quantities'. Example quantities include mass, time, speed, distance, work, energy etc. * Vector Quantities: Physical quantities which possess magnitude as well as direction are called ‘Vector Quantities’. Examples include Force, displacement, velocity, momentum, acceleration etc. ### Distance and Displacement * Distance: Total path covered * Displacement: It is the shortest path covered from initial to final position in a fixed direction. A diagram shows a square path with a start and end point marked as C. The square path consists of 2 right angles, labelled M. ### Distance and Displacement Numerical Examples A variety of examples are given * Example 1: a right angle triangle which has side lengths of 4km and 3km. The unknown hypotenuse is labelled $x = 5km$ * Distance = 3 + 4 = 7km * $Displacement = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5km$ * Example 2: Simple linear path which has side lengths of 3km and 4km, resulting in a total distance of 7km: * Distance = 3 + 4 = 7km * Displacement = 7km * Example 3: A linear path is shown which consists of two side lengths, 4km and 3km: * 4km Distance = 4 + 3 = 7km * Displacement = 1 km ### Difference between Distance and Displacement | Distance | Displacement | | :----------------------------------------- | :---------------------------------------------------------- | | Distance is the total path travelled. | The shortest distance measured from start to end point. | | It is a scalar quantity. | It is a vector quantity. | | It is either equal to or greater than displacement. | It is either equal to or less than distance. | ### Speed and Velocity * Speed ($Scalar$): $Speed = \frac{Distance}{time}$, SI unit of speed is $m/s$ * Velocity ($vector$): The rate of displacement travelled is called velocity, $Velocity = \frac{Displacement}{time}$ ### Difference between Speed and Velocity. | Speed | Velocity | | :------------------------------------------ | :------------------------------------------------------ | | The distance travelled in unit time. | The displacement of a body in unit time. | | $Speed = \frac{distance}{time}$ | $Velocity = \frac{displacement}{time}$ | | It is a scalar quantity | It is a vector quantity. | | It is either equal to or greater than velocity. | It is either equal to or less than speed. | ### Average Speed and Average Velocity Average Speed: The ratio of total distance to the total time is called average speed. $average \ speed = \frac{Total \ Distance}{Total \ Time}$ Average Velocity: The ratio of total displacement to the total time. $av = \frac{total \ displacement}{total \ time}$ Average velocity of a car is shown, and is calculated as: $Average \ velocity = \frac{v + u}{2}$ ### Acceleration and its types * Acceleration: $a = \frac{change \ in \ velocity}{time}$ * SI unit of aceleration is $m/s^2$ or $ms^{-2}$ * $a = \frac{V-u}{t}$ #### Types of Acceleration * Positive acceleration: Velocity is increasing * Negative acceleration/ Retardation/ Deceleration: Velocity is decreasing * Zero Acceleration: Velocity is constant Diagrams are provided to demonstrate this ##### Negative acceleration example A truck is shown with Force and acceleration acting in the opposite direction to velocity and displacement. ### Difference between Uniform Motion and Non-uniform Motion | Uniform Motion | Non-Uniform Motion | | :-------------------------------------------------------------------------------- | :----------------------------------------------------------------------------------- | | The motion in which the object covers equal distances in equal intervals of time. | The motion in which the object covers unequal distances in equal intervals of time. | | In uniform motion acceleration is zero. | In non-uniform motion acceleration is not zero. | #### Numerical examples * Uniform Motion: an object covers 10m every second * Non-Uniform Motion: an object covers 10m, then 12m, then 18m, then 32m every second ### Graphs Various distance versus time graphs for motion are shown. * **Graph 1**: In rest. The horizontal axis is labelled time and the vertical axis is labelled Distance. * **Graph 2**: Uniform speed or constant speed. The horizontal axis is labelled time and the vertical axis is labelled Distance. The gradient is equal the speed ($Slope = \frac{Distance}{time} = Speed$). * **Graph 3**: Constant velocity. The horizontal axis is labelled time *t*, and the vertical axis is labelled *velocity*. * Area gives Displacement, $v * t = Displacement$ * **Graph 4**: Non constant velocity: * AB = uniform retardation * BC = uniform acceleration * **Graph 5**: Time vs velocity. The horizontal axis is labelled time *t*, and the vertical axis is labelled *velocity*. The slope gives acceleration. The area gives displacement. * $Slope= \frac{{V_2} - {V_1}}{t}$, $a = \frac{{V_2} - {V_1}}{t}$ * **Graph 6**: Velocity versus time. * initial velocity = 0 * AB = uniform acceleration * BC = constant velocity * CD = unifrom retardation An additional 2 graphs are provided in order to asses understanding. * **Graph 7**: * A) constant speed B) non-uniform motion C) Zero acceleration D) None * Answer B) non-uniform motion * **Graph 8**: * A) uniform acceleration B) non-uniform acceleration C) constant velocity D) None * Non uniform acceleration ### Equation of Motion $V =$ final velocity $u =$ initial velocity $a =$ acceleration $s =$ displacement $t =$ time * $v = u + at$ * $s = ut + \frac{1}{2}at^2$ * $v^2 - u^2 = 2as$ #### Proofs of the equations of motion * $First: V= u + at$ * $Second: s = ut + \frac{1}{2}at^2$ * $Third: 2as = v^2 - u^2$ ### Free fall motion * acceleration due to gravity = $9.8 m/s^2 \approx 10 m/s^2$ * $g = a = 10m/s^2$ Diagram is shown with Earth, and $force \ due \ to \ gravity$. ### Uniform circular motion * When an object moves in a circular path with uniform speed, its motion is called uniform circular motion. $T = \frac{Distance}{Velocity}$ * $Time \ taken \ to \ complete \ one \ revolution = Time \ Period$ * $T = \frac{2{\pi}R}{V}$ A circle is drawn, where r is the radius and s is the circumference. ## Questions * When in motion the body changes its _____ with respect to its surroundings. * A) Size B) Shape C) Position D) No change * Correct answer C) Position * In the given figure, object starts its motion at point A, goes to B-C-D and again came back to A. Find displacement from point A to C. * A) 0 B) 16 C) can't say D) 5m * Correct answer D) 5m A rectangle diagram is shown with lengths 3m and 4m. AC is the diagonal length and is shown to be calculated with pythagoras theorum. * When an athlete runs on a circular track and reaches the same point _____. * A) Displacement = distance B) Distance > displacement C) Displacement > 0 D) Distance < displacement * Correct answer B) Distance > displacement A circle is draw where Displacement = 0 and Distance = 2π R * A cheetah is the fastest land animal and can achieve a peak velocity of 100 km/h up to distance less than 500 m. If a cheetah spots its prey at a distance of 100 m, what is the minimum time it will take to get its prey, if the average velocity attained by it is 90 km/h? * A) 2 sec B) 3 sec C) 4 sec D) None * Correct answer C) 4 sec *Derivation* $As = 90 km/h = 90 × \frac{5}{18} m/s$ $As = 25 m/s$ $time = \frac{100}{25} = 4 sec$ * An object travels 16 m in 4 s and then another 16 m in 2 s. What is the average speed of the object? * A) 6 m/s B) 8 m/s C) 4 m/s D) None * Correct answer = 16/3 m/s *Derivation* $As = \frac{Total \ Distance}{Total \ time}= \frac{16 + 16}{4 + 2} = \frac{32}{6}= \frac{16}{3} m/s$ * Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha. * A) 3 m/s, 0 B) 0, 3 m/s C) 6 m/s, 3 m/s D) None * Correct answer 3/s, 0 *Derivation* $\frac{Total \ Distance}{Total \ time} = \frac{180}{6} = 3 m/s$ Average velocity = 0 A bus decreases its speed from 80 km h-1 to 60 km h-1 in 5 s. Find the acceleration of the bus. *(HW)* $u= 80 km/h, v = 60 km/h, t= 5 s$ $ a = \frac{v - u}{t} =(60 \times \frac{ 5}{18} - 80 \times \frac{ 5}{18}) /5$ $a= -1.1 m/s^2$ A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h-¹ in 10 minutes. Find its acceleration *(HW)* A trolley, while going down an inclined plane, has an acceleration of 2 cm s-2. What will be its velocity 3 s after the start? *(HW)* A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s² for 8 s. How far does the boat travel during this time? *(HW)* A odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of the trip. If the trip took 8 h, calculate the average speed of the car in km h-1 *(HW)* Abdoul while driving to school, computes the average speed for his trip to be 20 km h-1. On his return trip along the same route, there is less traffic and the average speed is 30 km h-1. What is the average speed for Abdul's trip? *(HW)* A car accelerates uniformly from 18 km h-¹ to 36 km h-¹ in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time *(HW)*. A bus starting from rest moves with a uniform acceleration of 0.1 ms-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled *(HW)*. A racing car has a uniform acceleration of 4 ms-2. What distance will it cover in 10 s after start? *(HW)* The brakes applied to a car produce an acceleration of 6 ms-2 in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time *(HW)*. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth *(HW)*. Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 ms-¹ in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 ms-¹ in the next 5 s. Calculate the acceleration of the bicycle in both the cases *HW*. $acceleration (a) = \frac{change in velocity}{time}$ $a = \frac{4-6}{5}= -0.4ms^{-2}$ A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms-2, with what velocity will it strike the ground? After what time will it strike the ground? *(HW)* A stone is thrown in a vertically upward direction with a velocity of 5 ms-¹. If the acceleration of the stone during its motion is 10 ms-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there? *(HW)* A train is travelling at a speed of 90 km h-1. Brakes are applied so as to produce a uniform acceleration of -0.5 ms-2. Find how far the train will go before it is brought to rest *(HW)*. Four cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown in figure. Which are is the slowest? *(HW)* What can you say about the motion of an object if its speed-time graph is a straight line parallel to time axis *(HW)*? What is the shape of velocity-time graph for non-uniformly accelerated motion? *(HW)* Draw the distance-time graph for a car moving with non-uniform speed *(HW)*. State with an example which of the following situations are possible *(HW)*: (a) An object with a constant acceleration and zero velocity. (b) An object moving with acceleration but with uniform speed. (c) An object moving in a particular direction with acceleration in the perpendicular direction? (1) A situation where the object is at rest and in motion simultaneously. (2) A motion in which acceleration is non-uniform. (3) A motion in which acceleration is in the direction of motion of an object. (4) An example where an object moves in a certain direction with an acceleration in the perpendicular direction. (5) An object with constant acceleration and zero velocity. A train moves with a speed of 30 km/h in the first 15 minutes, with another speed of 40 km/h in the next 15 minutes, and then with a speed of 60 km/h in the last 30 minutes. Calculate the average speed of the train for this journey *(HW)*. The velocity-time graph of SUV is given below. The mass of the SUV is 1000 kg. (a) What is the distance travelled by the SUV is first 2 seconds? (b) What is the braking force at the end of 5 seconds to bring the SUV to a stop within one seconds? *(HW)* A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms-2, with what velocity will it strike the ground? After what time will it strike the ground? *(HW)* $ ut + 1/2 a t^2(s = ut + 1/2 at^2)$ Abdul, while driving to school, computes the average speed for his trip to be 20 km h-1. On his return trip along the same route, there is less traffic and the average speed is 30 km h-1. What is the average speed for Abdul's trip? *(HW)* An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of minutes 20 s. *(HW)* ## Summary $a= \frac{v-u}{t}$ * $v=u+at$ * $s=ut+\frac{1}{2}at^2$ * $v^2-u^2=2as$ snth $s_{nth}=u + \frac{a}{2}(2n-1)$ $T=\frac{2 \prod R}{V}$ ### Graphical derivation Slope = acceleration $= \frac{v-u}{t}$ #THANK \## YOU Let me know if there are any changes.