kinematics class 11 ncert chapters
CBSE Class 11 Syllabus 2017-2018: All Subjects. Physics and Measurement. T + $\frac{1}{2}$(–g) T2, Or, $\frac{1}{2}$g T2 = u sin$\theta $. Let us consider two objects A and B moving with velocities at $\overrightarrow{{{V}_{A}}}$and $\overrightarrow{{{V}_{B}}}$an angle ‘$\theta $’. (C) Distance :The length of the actual path traversed by the particle is termed as its distance.Distance = length of path ACB. The initial vertical distance between points of projection is 30 - 10 = 20 m. This vertical distance is covered with component of relative velocity in vertical direction. = velocity of the man with respect to the ground. So maximum seperation = S1 - S2 = 400 - 200 = 200 m* When they meet both will travel the same distance. You simply subtracted my wife's height from my height. (IV) Drawing of graphs on the basis of given information :(a) If acceleration of the body is zero : (i) If the velocity of the body is v0 and it starts from origin : (b) If a body has constant acceleration : (i) u0, x0 & a0 are positive constants. Physics With AJ is an educational website related to physics, created by a physics lecturer to help students with physics. Let us discuss both the cases separately. of Kinematics, Chapter Notes, Class 11, Physics (IIT-JEE & AIPMT) | EduRev Notes for JEE, the answers and examples explain the meaning of chapter in the best manner. To find the relative velocity of ‘A’ w.r.t. Here s is taken negative because it is in the opposite direction of initial velocity. or y = px - qx2, where p and q are constants. We should ensure that collision does occur at the point of return. NCERT Books for Class 11: If you are looking for NCERT books for Class 11, you have come to the right place.NCERT books are important resources for your exam preparation. Read and download free pdf of CBSE Class 11 Physics Kinematics Assignment. EduRev is like a wikipedia
another object. For uniform velocity, acceleration is zero. Thus, Range (R) or the distance OB is also equal to be displacement of projectile along x-direction in the. Suppose that two particles are projected from the ground with speeds u1 and u2 at angles α1 and α2 as shown in Fig.A and B. or h = 58.8 mEx. Hence, two angles of projection for the same horizontal range are $\theta $ and (90 –$\theta $ ). (i) How high will the ball rise and(ii) how long will it be before the ball hits the ground?Sol. In order to measure physical quantities, a certain basic, internationally accepted reference standard is … $\theta $= 0). The speed of an object calculated for very small interval of time is called instantaneous speed. If the projectiles collide in the air, then find the distance "s" between the towers. Here, negative sign implies that velocity of particle at Q is along negative y direction. 49: Two ships A and B are 10 km apart on a line running south to north. Distance travelled by car in this time (20 sec). Chapter 4. * Rest and motion are relative terms. Let $\overrightarrow{V}$ be the velocity of the projectile at any instant. (iv) Average velocity = Ex. 21: A balloon is ascending at the rate of 9.8 ms-1 at a height of 39.2 m above the ground when a food packet is dropped from the balloon. 28: If at t = 0 u = 5 ms-1 then velocity at t = 10 sec, = 55 ms-1Ex. This is required expression for time of flight. your solution of Kinematics, Chapter Notes, Class 11, Physics (IIT-JEE & AIPMT) | EduRev Notes search giving you solved answers for the same. Class 11 Physics Notes Chapter 1 Physics World And Measurement PDF. 40: A ball rolls off the top of a stairway with a constant horizontal velocity u. Now, let us take one example of this.Ex. Then, (a) At point Q, x-component of velocity is zero. (v) A particle starts from x = 0 and initial speed 10 ms-1 and moves with constant speed 10ms-1 for 20 sec. $\therefore $ Snth = u + $\frac{\text{a}}{\text{2}}$ (2n – 1)…………………is the required expression. Gravitation. vinstantaneous = Slope of the tangent x - t curve(iii) Reading of Graph :(A) Reading x v/s t graphs Explanation : (2) Body starts from origin and is moving with speed tan q away from origin. The branch of physics which deals with the study of motion as the function of time is known as kinematics. Take g = 10 m/s2 Sol. If their are two vectors and and their resultent make an anlge a with and b with . Its unit is metre (m). You can also find Kinematics, Chapter Notes, Class 11, Physics (IIT-JEE & AIPMT) | EduRev Notes ppt and other JEE slides as well. Let us consider a body moving initially with velocity ‘u’ is accelerated to distances Sn and Sn-1 in ‘n’ sec and (n-1) sec respectively. 54: An aircraft flies at 400 km/h in still air. Reaction time depends on complexity of the situation and on an individiual.One can measure one's reaction time by a simple experiment. 24: Draw the graph for the equation : 2y + 4x + 2 = 0Sol. From t = t1 to t2 acceleration is negative, â displacement = area under the v - t curve.Ex. Your email address will not be published. (d) Time to reach half of the maximum height : Equation 1 gives two value of time which corresponds to, t1 = (from ground to Hmax/2 in upward motion). Let $\overrightarrow{V}$ be the velocity of the projectile at any instant. Find the direction he must steer and time of his journey if AB = 1000 km.Sol. The total distance covered is 600 m. Find the acceleration, retardation and the total time taken.Sol. Now, let us derive the expressions for time of flight (T) and range (R) along the plane. The ball attains a maximum height Hmax. (ii) Derivation of S = ut + $\frac{\text{1}}{\text{2}}$ at2 for a uniformly accelerating body by graphical method. As Dt tends to zero, the ratio defining velocity becomes finite and equals to the first derivative of the position vector. NCERT Exemplar Class 11 Physics book published by NCERT (National Council of Education and Research) is a … Here, particle "A" follows "B", "B" follows "C" and "C" follows "A". Taking horizontal component, equation (vi) becomes, Or, Vx = ucos$\theta $ {$\because $ ax = 0}. Chapter 8. (ii) speed of rain is, = Ans.Relative Motion between Two Projectiles :Let us now discuss the relative motion between two projectiles or the path observed by one projectile of the other. (b) Now, = (30 + 50 cosα)m/s = = (30+ 40) m/s. (v) For t < t1 velocity of A is greater than velocity of B. MOTION IN A PLANE part-1. Let ‘Vx’ and ‘Vy’ be the components of velocity $\overrightarrow{V}$along x-axis and y-axis respectively and ‘$\alpha $’ be the angle made by the velocity with the horizontal. v = tanq = = (G) Instantaneous velocity :Instantaneous velocity is defined exactly like speed. Then. When time = t, displacement travelled = s (say). (2) Acceleration of the body is constant and positive. Its velocity is decreasing with time and at t = t0 . The distance travelled d and the reaction time tr are related by, Note :Definition : Time taken by a driver to react for a situation, Reaction Time of the driver is Ît = t1 - t0Total distance covered by the car before stopping, = distance covered in uniform motion during to to t1 + distance, cover in deaccelerated motion during t1 to t2 = Total distance = u(Ît) + Ex. Distance covered by a body in nth second ( Snth formula): The distance covered by a body in the interval of ‘n’ sec and (n-1) sec is called Snth formula. 16: What is the maximum separation between car and scooter ?Sol. Draw perpendicular BM to t-axis AN perpendicular to BM. (5)Body starts from x = x0 and is moving toward the origin with constant velocity passes throw origin after same time and continues to move away from origin. s = = or s = 62.64 m Ans.Ex. Our Class 11 Physics revision Notes of Motion in a Plane cover the basics of motion in two dimensions. River is flowing along positive x-direction with velocity . (i) Find at what distance will the bus overtake the car ? ... Kinematics Unit 3: Laws of Motion Unit 4: Work, Energy and Power Draw BM perpendicular to t-axis and AN perpendicular to BM. First find out it what instant velocity of block becomes zero. Let us consider a body initially at So is moving with constant velocity for time ‘t’ where it covers distance ‘S’. Its acceleration is increasing for whole of its motion. (iii) Velocity of A is greater than that of B. (i) Derive v = u + at for a body with constant acceleration by graphical method. Assume that effect of air friction and wind resistance are negligible and value of `acceleration due to gravity is constant. (3) Acceleration of the body is constant and negative, (4) Initially the acceleration of the body is zero. It is a two dimensional, two body problem with zero acceleration. The rate of change of displacement of an object with respect to another object when both are in motion is known as relative velocity of one object w.r.t. = 3 km/h cos 30° + 2 km/h cos 90° = km/h, Displacement along the X-axis as the man crosses the river, = à = Ex. this is your one stop solution. Finally its motion ends at t = t1 at x = x2 m. (ii) Body B starts its motion at t = 0 from x = x0 and then moves with constant velocity away from the origin. Reaction time is the time a person takes to observe, think and act. A general quadratic equation represents a parabola. www.ncerthelp.com (Visit for all ncert solutions in text and videos, CBSE syllabus, note and many more) Physics Notes Class 11 CHAPTER 3 MOTION IN A STRAIGHT LINE Motion If an object changes its position with respect to its surroundings with time, then it is called in motion. It is the total time spent by projectile in air. It is defined as the ratio of total displacement covered to total time taken. (b) Find out its velocity at x = 3 mSol. This is
Now, let us derive some standard results and their special cases. ax = 0, ay = - g, vx = u cosθ, and vy = u sinθ - gt, and, x = u cosθ à t y = u sinθ t - gt2. At the highest point, speed of the particle is minimum. (b) At what point on the opposite bank will he arrive ?Sol. The front windscreen of a moving car gets wet in rain while the behind screen remains dry. 4: Velocity-time equation of a particle moving in a straight line is. Hence, to reach point B the boatman should row at an angle θ= upstream from AB. Taking vertical component, equation (iii) becomes, $\therefore $ 0 = u2sin2$\theta $ + 2 (-g) Hmax, $\therefore $ Hmax = $\frac{{{u}^{2}}si{{n}^{2}}\theta }{2g}$. It can be assumed that B is at rest and A is moving with . (iii) Derivation of v2 = u2 + 2as for a uniformly accelerating body by graphical method. Calculate the overall stopping distance of the car for a speed of 60 km-1 of the car.Sol. Let T be its time of flight. Q. The acceleration along x–axis is zero (ax = 0). Let us find. (B) Position or location : It is defined with respect to some reference point (origin) of given frame of reference.Consider a particle which moves from location (at time t1) to location (at time t2) as shown in the figure below. Similarly taking vertical component, equation (v) becomes, V = $\sqrt{{{V}_{x}}^{2}+{{V}_{y}}^{2}}$= $\sqrt{{{u}^{2}}+{{g}^{2}}{{t}^{2}}}$, Or, V = $\sqrt{{{u}^{2}}+{{g}^{2}}{{t}^{2}}}$, And direction, $\tan \alpha $= $\frac{{{V}_{y}}}{{{V}_{x}}}$, Or, $\alpha $= tan –1$\left( \frac{gt}{u} \right)$, For the velocity with which the projectile hits the ground, we replace the time ‘t’ by time of flight ‘T’ and angle ($\alpha $) by the striking angle ($\beta $) then magnitude and direction becomes, Magnitude, V =$\sqrt{{{u}^{2}}+{{g}^{2}}{{T}^{2}}}$, And direction, $\beta $= tan –1$\left( \frac{gT}{u} \right)$, $\theta $ = sin–1$\left( \frac{{{V}_{R}}}{{{V}_{S}}} \right)$, (b) Velocity along AB, VAB = Vs cos$\theta $, (c) Time to cross the river (t) = $\frac{AB}{Vscos\theta }$, Time to cross the river, t = $\frac{AB}{{{V}_{s}}cos\theta }$, Tan$\theta $ = $\frac{{{V}_{R}}}{{{V}_{s}}}$= $\frac{BC}{AB}$, $\therefore $Distance apart from the opposite end (BC) = $\frac{{{V}_{R}}}{{{V}_{s}}}$× width of river {OR, BC Tan$\theta $ × width of river}. Why? Given a = 2x. If he is crossing a river where the current is 2 km/h. 19: A ball is thrown vertically upwards with a velocity of 20 ms-1 from the top of a multi-storeyed building. e.g. Take g = 9.8 ms-2.Sol. SYSTEM OF PARTICLES AND ROTATIONAL MOTION . The distance travelled by car in 10 sec. t, For vertical motion : u = u sinθ, a = -g, so the vertical distance covered in time t is given by, s = ut + or y = u sin θ.- or y = x tanq - ...(1). We see here that projectiles are approaching both horizontally and vertically. (iv) A t = t2, Both A & B are at the same distance from starting point that means B overtakes A at t = t2, (v) Q velocity of both A & B are constant. To download notes, click here NOW: http://bit.ly/2wlXDOFUnacademy JEE brings you another Physics session to prepare you for JEE Mains 2020. It then runs at constant velocity and finally brought to rest in 200 m with a constant retardation. Here, it is important to note that is the velocity of boatman with which he steers and is the actual velocity of boatman relative to ground. The chapters present in the NCERT Solutions for Class 11 Physics are – 1 – Physical World 2 – Units and Measurements 3 – Motion in a Straight Line 4 – Motion in a Plane 5 – Laws of Motion 6 – Work, Energy, and Power 7 – Systems of Particles and Rotational Motion 8 – Gravitation 9 – Mechanical Properties of Solids 10 – Mechanical Properties of Fluids 11 – Thermal Properties of Matter 12 – … This online service offers easy access to the NCERT textbooks. Then find out the time and displacement at which ball have half of the maximum speed.Sol. Net distance for which body moves with uniform velocity, Total time of journey, t = (20 + 10 + 20) sec, Average velocity = = 12 m/s. ux = u cos$\theta $ and uy = u sin$\theta $ along horizontal and vertical direction respectively. In other words, the component velocities in x-direction should be equal to that two projetiles cover equal horizontal distance at any given time. The total time of a projectile spent in air is known as time of flight. ux = Vx. Relative Motion :The word 'relative' is a very general term, which can be applied to physical, nonphysical, scalar or vector quantities. (ii) The magnitude of velocity is equal to speed. Hence, to reach the point directly opposite to starting point he should head the boat at an angle of 30° with AB or 90° + 30° = 120° with the river flow. (e.g-mass, length, volume, density)Vector Quantities:-Vector quantities are those quantities which require magnitude as well as direction for their complete specification. (a) In what direction will his boat be headed, if he wants to reach a point on the other bank, directly opposite to starting point? For example, if a person is driving and suddenly a boy appears on the road, then the time elapse before he applies the breaks of the car is the reaction time. Motion under gravity :I format (When a body is thrown vertically upward) :It includes two types of motion. Important topics of 11th Physics are covered. â t = 0 and So, T = Range (R) : The horizontal distance covered by the projectile during its motion is said to be range of the projectile. (ii) u º = initial velocity. Acceleration vs time graph velocity vs time graph, position vs time graph(V) Conversion of velocity v/s time graph to speed v/s time graph.As we know that magnitude of velocity represent speed therefore whenever velocity goes -ve take its mirror image about time axis.Ex. from t = 0 to t = t1, x - t graph will be a straight line. When a particle is dropped then it will automatically attains the velocity of the frame at that time.Ex. Class 11 Physics Notes Chapter 2 Kinematics PDF. Its SI unit is m/s2. vy = u sin θ - gt, and y = u sin θ t - Ex. It is equal to OA = R. Thus R = Horizontal velocity à time of flight = u à T. or R = uVelocity of the Projectile at any Instant : At the instant t (when the body is at point p), let the velocity of the projectile be v. The velocity v has two rectangular components: Vertical component of velocity, vy = 0 + gt = gt, If the velocity v makes an angle b with the horizontal, then.