Physics

Session Kinematics - 2

Session Opener REST ! or MOTION

Session Objectives

Session Objective 1.Rest and motion 2.Distance and displacement 3.Uniform and non-uniform motion 4.Velocity 5.Acceleration 6.Equations of motion

Rest and Motion Body at rest : Position constant w.r.t. fixed point as time increases Fixed Point : Origin of a coordinate system Y O 90 0 x X r y P (position) (origin) 

What is motion ? Change in position of an object with time, with respect to a given co ordinate system. Motion Actual distance traveled : Curve P 0 P 1 P 2 P 3 P 4. Displacement : Straight line P 0 P 4 directed from P 0 to P 4 P 0 (t=0) P 1 (t=t 1 ) P 2 (t=t 2 ) P 3 (t=t 3 ) P 4 (t=t 4 ) x y

Distance and Displacement 1. Distance  Displacement. 1 2 x t o 2. Distance = Displacement (If direction remains same.) 3.Distance is always 0 or+ve. Displacement can be +ve,0 or –ve. 4. Distance always increases with motion.

Average Speed P 0 (t=0) P 1 (t=t 1 ) P 2 (t=t 2 ) P 3 (t=t 3 ) P 4 (t=t 4 ) x y

Courtesy:

Instantaneous Speed When time is infinitesimal (=t  0), distance is infinitesimal(=s  0) Instantaneous speed = tt ss tt t ss PoPo P1P1 s Speed : Scalar Unit m/s Dimension LT -1

Average Velocity Average velocity is defined as displacement divided by time taken. Nature : vector Dimension : [LT -1 ] Unit : m/s Displacement and average velocity are in same direction

Instantaneous Velocity If t= (t 2 – t 1 )is extremely small (t  0) v is instantaneous velocity v is a vector. Unit of v : m/s

Uniform Motion (One dimension) Equal displacement (x) traveled in equal time interval (t) v is constant. x = x 0 + vt If x 0 = 0 at t = 0, x = vt x0x0 x1x1 x2x2 x3x3 x4x4 x 0 t1t1 t2t2 t3t3 t4t4 t

Class Exercise

Class Exercise - 3 Graph in the figure below shows the variation of displacement with respect to time for a particle in one-dimensional motion. Which of the following represents the velocity-time graph of the particle in motion? Options is in next slide

Class Exercise - 3 (a) (b) (c) (d)

Solution - 3 For t = 0 to 5 Hence answer is (c)

Class Exercise - 8 An object travels half the distance with v 1, with v 2 for half of the remaining time and with v 3 for the remaining half of the time. If the object never reverses the direction of motion, find the average velocity during the motion. Solution :

Average Acceleration Change in velocity divided by the time interval during which the change occurs. Nature : vector unit : m/s² Dimension: [LT -2 ]

Non uniform motion (constant acceleration) Instantaneous acceleration t v v u t1t1 t2t2 t=t 2 -t 1 For constant acceleration v = u + at Equation of motion (1)

Class Exercise

Class Exercise - 1 Which graph represents increasing acceleration? (a) A(b) B(c) C(d) None of these

Solution - 1 Hence answer is (b). Increasing a  Increasing slope of v-t curve. By observation, we find that the velocity is increasing at an increasing rate. So acceleration is increasing.

Class Exercise - 9 An object starts from rest. It accelerates at 2 m/s 2 till it reaches its maximum velocity. Then it retards at 4 m/s 2 and finally comes to rest. If the total time taken is 6s, find v max and the displacement of the object.

Solution - 9 V max = 2t 1 V max = 4t 2 s = Area of triangle  t 1 = 4 s, t 2 = 2 s Hence, V max = 8 m/s

Class Exercise - 2 A particle is thrown vertically upwards with velocity v. It returns to the ground in time T. Which of the following graphs correctly represents the motion? (a) (b) (d) (c)

Solution - 2 The acceleration is constant (= –g). So slope has to be negative throughout the motion and velocity varies between v and –v. Hence answer is (c).

Class Exercise - 5 If a particle has an initial velocity of and an acceleration of, its speed after 10 s is (a) 10 units(b) 7 units (c) units(d) 8.5 units

Solution - 5 Hence answer is (c).

One dimensional equations of motion Distance s = area under v-t graph = ½ (u+v)t But, s = v avg t Hence, v avg = (u+v)/2 t v v u t1t1 t2t2 t=t 2 -t 1 Using equation of motion (1) s = ut + ½ at 2 Equation of motion (2) Equation of motion (3)

One dimensional equations of motion v = u + at v² = u² + 2as Distance traveled in n th second

Class Exercise

Class Exercise - 6 A particle moves along the X-axis as x = u(t – 3) + a(t – 3) 2, then Which of the following are true? (a)initial velocity of particle is u at t = 0 (b) acceleration of particle is a (c) at t = 3 the particle was at origin (d) the particle may have negative velocity

Solution - 6 The observation of displacement has started at time t = 3 s, after the object has actually started. So if it represents the time for which the object has traveled and s be the displacement after the observation has started, then general form is

Class Exercise - 10 The magnitude of maximum acceleration, retardation of an object is ‘a’ m/s 2. What is the minimum time taken by the object to cover a displacement ‘s’ if it starts from rest and finally comes to rest? Solution : The minimum time would be when the acceleration is at maximum and deceleration is also maximum. Half the time accelerating at a and the rest of the time deceleration at ‘a’.

Uniform and Non-Uniform Motion Let us see a comparison of uniform and non-uniform motion Courtesy:

Class Exercise

Class Exercise - 7 An object has one-dimensional motion. If V = 6t + 4t 3, then (a) what is the distance covered from t = 3 s to t = 5 s? (b) when is the acceleration < 0 for the first time? So acceleration is never negative. Solution :

Class Exercise - 4 A particle moves in a straight line, starting from rest. The acceleration of the particle is given by What is the distance traveled by the particle in the time interval 0 to  seconds.

Solution - 4 [Check that v(0) = 0]

Solution - 4  x = 2 – log( + 1)

Thank you