Motion in a straight line class 11 notes
A body is said to be in motion when it changes its position with respect to the observer while it is said to be at rest when there is no change in its position with respect to the observer. For instance, two passengers travelling in a moving train are at rest with respect to each other but in motion for a ground observer.
Physically, a particle is considered as analogues to a point. A body with a definite size is considered as a particle when all of its parts have the same displacement, velocity and acceleration. The motion of any such body can be studied by the motion of any point on that body.
Position and Position Vectors
The position of any point can be represented by its coordinates with respect to an origin in the Cartesian system. For example, point A can be represented by (xA, yA, zA).
Given the coordinates of two points A & B, the position vector of B w.r.t. A can be determined as follows:
Distance and Displacement
A particle follows either a curve or a straight line when moving in the space. This curve or line is known as its trajectory. Distance is the length of the path or trajectory covered by the particle and displacement is the difference between the vectors of the first and the last position on this path.
Distance and displacement are illustrated in the fig. where AB (curve length) is the distance and vector ∆ṝAB is the displacement (ṝB – ṝA).
Following points should be considered about both the quantities:
(a) Distance is a scalar quantity while displacement is a vector.
(b) Displacement is always less than or equal to the distance in magnitude.
(c) Displacement can be zero but distance cannot be zero for a moving body.
Average Speed and Average Velocity
The total distance travelled by a particle divided by the total time taken is known as its average speed.
Average speed = Total distance travelled/Total time taken
While the average velocity is defined as vav = displacement/time elapsed
Or v = rB – rA/tB – tA = ∆rAB/∆t
Both average speed and average velocity are expressed in ms-1 or kmh-1, the former is a scalar quantity while the latter is a vector.
INSTANTANEOUS SPEED AND INSTANTANEOUS VELOCITY
The average speed/velocity of a moving body with an infinitesimally small time interval (i.e. ∆t → 0) is known as instantaneous speed/velocity. Therefore
(a) As the time interval tends to zero (i.e. ∆t → 0), the displacement vector ∆ ṝ is along the direction of motion of the particle i.e. tangential to the path of the particle at that instant. Thus, the instantaneous velocity direction is always tangential to the path of the particle.
(b) Instantaneous speed and the magnitude of instantaneous velocity are always the same. i.e. Instantaneous Speed= | Instantaneous Velocity |
A particle moving on a straight line, say along the x-axis, has an instantaneous velocity as follows
A particle is said to move in a uniform velocity when the velocity of a particle remains constant with respect to time. It is said to be accelerated when velocity changes with respect to time.
Average acceleration and Instantaneous Acceleration
The rate of change of velocity is defined as acceleration. Velocity changes with the change in magnitude or direction or both.
(a) Acceleration is a vector quantity and its SI unit is ms-2.
(b) The average acceleration vector and the change in velocity vector are in the same direction.
(c) The direction of the velocity vector and the direction of the acceleration vector are independent of each other.
(d) Acceleration is perpendicular to the velocity vector only when there is a change in direction of velocity with time, with its magnitude being constant.
If a body moves with uniform acceleration along a straight line, then the average acceleration and instantaneous acceleration will always be the same.
UNIFORMLY ACCELERATED MOTION FOR 1-D MOTION
Equations of Motion
Consider that the acceleration of a particle ‘a’ is constant.
First Equation: Acceleration is defned as the rate of change of velocity
a = dv/dt => dv = a.dt … (i)
The velocity at time 0 is u and at time t is v. Thus, at t = 0, v= u and At t = t, v = v.
Integrating equation (i) for these limits: V = v to u = v and t = 0 to t = t
Second Equation: Velocity is defned as the rate of change of displacement.
v = ds/dt => ds = vdt; ds = (u + at)dt [∴v = u + at] … (ii)
Suppose the position of the particle is ‘0’ at time ‘0’ and ‘s’ at time ‘t’. Hence, at t = 0, s = 0 and at t = t, s = s
Integrating equation (ii) for these limits:
Third Equation: From the defnition of acceleration
a = dv/dt => a= dv/dx . dx/dt; a = dv/dx .v => vdv = adx … (iii)
If the velocity at position 0 is u and at position, s is v, at s=0, v=u and at s=s, v=v
Integrating equation (iii) for these limits: V=u to v = v and x = 0 to x = s
Displacement In nth Second
Displacements Sn of a particle in n seconds and Sn-1 in (n-1) seconds are given as: sn = un + ½ an2
sn – 1 = u(n – 1) + ½ a(n – 1)2
Displacement in nth second: sn – sn-1 = un + ½ an2 – u(n – 1) + ½ a(n – 1)2 = u + a/2 (2n – 1)
MOTION OF BODY UNDER GRAVITY (FREE FALL)
The force of attraction that the earth exerts on all the bodies is called the force of gravity and the acceleration induced by gravity is called acceleration due to gravity, represented by g. All bodies irrespective of their size, weight or composition fall with the same acceleration near the surface of the earth in the absence of air. The motion of a body falling towards the earth from a small altitude (h<<R) is known as free fall (R: Radius of Earth).
Body Projected Vertically Upwards
(a) Equation of motion: Considering the point of projection as origin and direction of motion (i.e. vertically up) as positive a = – g [as acceleration is downwards]
If a body is projected with velocity u and after time t it reaches to a height h then
v = u – gt … (i)
h = ut – ½ gt2 … (ii)
v2 = u2-2gh … (iii)
(b) For maximum height (H): v =0
Using equation (ii) we get 02 = u2 – 2gH => H = u2/2g (c) Time taken to reach maximum height (t): v=0
Using equation (ii) we get: 0 = u-gt; T = u/g.
(c) Time of ﬂight (T) is the time during which the object travels. In this case, it is the time between the maximum height and the ground.
Thus, h = 0. Using equation (iii): 0 = uT – ½ gt2; 0 = T(u – ½ gt)
⇒ either T=0 or T=2u/g = 2 x. Time taken to reach maximum height (t)
(d) The following graphs show the displacement, velocity and acceleration with respect to time (for maximum height) when the body is thrown upwards:
(a)Time taken by the body to travel up is equal to the time taken by the body to fall down. Time of descent (t2) = time of ascent (t1) = u/g
(b) The speed with which a body is projected up is equal to the speed with which it comes down. The magnitude of velocity at any point is the same whether the body is moving up or down.
Body Dropped From Some Height (Initial Velocity Zero)
(a) Equations of motion: Let the initial position be the origin and direction of motion (i.e. downward direction) to be positive, then
u = 0 [As body starts from rest] a = g [As acceleration is in the direction of motion] Therefore, equations of motion are:
v = u + gt …(i)
v = ut + ½ gt2 …(ii)
v2 = u2 + 2gh …(iii)
(b) Velocity (v) of the particle just before hitting the ground: h = H. Therefore, using equation (iii): V2 = 0 + 2gH ; v = √2gH (c) Time (t) taken by the object to reach the ground: h = H
Therefore, using equation (ii): H = (0)T + ½ gT2; T = √(2H/g).
(c) The following graphs show the distance, velocity and acceleration with respect to time (for free fall):
(d) The distance covered in the nth sec, hn = ½ g (2n 1)
Hence the ratio of the distance covered in 1st, 2nd, 3rd sec, etc. is 1:3:5 i.e. only odd integers. These results obtained are the corollary of the Galileo’s Theorem:
For a uniform accelerating body, the distance travelled is always an odd ratio, i.e. 1:3:5:7, for a regular time interval.
(e) As h = (1/2) gt2, i.e. h α t2, distance covered in time t, 2t, 3t, etc., will be in the ratio of 12:22:32, i.e. square of integers.
Body Thrown Vertically from a Height
There are two possibilities when an object is thrown.
First, when an object is thrown in the upward direction, velocity is upwards whereas acceleration acts downwards, i.e. they are in opposite directions. Hence initially the object undergoes retardation and rises through a certain height and then it undergoes free fall from that height.
Second, when an object is thrown from a certain height, both the velocity and acceleration are in the downward direction, i.e. velocity and acceleration are in the same direction. In this case, the object undergoes acceleration. It hits the ground with a speed greater than the speed if it had gone through free fall.
Equations of motion are used in both cases. Assume a sign convention for the direction. Then, note the values of displacement, velocity and acceleration with appropriate signs. Finally, use the appropriate equations of motion.
NON-UNIFORMLY ACCELERATED MOTION
Equations of motion cannot be considered for particles travelling with constant or uniform acceleration. While deriving equations of motion, we considered acceleration to be constant. So, for solving non-uniformly accelerated motion, we will follow the two basic equations:
Vector quantity is not required for one-dimensional motion. Therefore the above equations can be re-written as:
(i) v = ds/dt (ii) a = dv/dt = v.dv/ds
With the displacement of a body plotted on the y-axis against time on the x-axis, displacement–time curve is obtained.
(a) The slope of the tangent at any point of time gives the instantaneous velocity at any given instant.
(b) At a uniform motion, the displacement–time graph is a straight line.
(i) If the graph obtained is parallel to the time axis, the velocity is zero.
(ii) If the graph is an oblique line, the velocity is constant.
Displacement, velocity and acceleration, specifying the entire motion, can be determined by the velocity-time curve.
(a) Instantaneous acceleration can be obtained by the slope of the tangent at any point corresponding to a particular time on the curve.
(b) Displacement during a time interval is obtained from the area enclosed by the velocity-time graph and time axis for a time interval.
(c) For a uniformly accelerated motion, the velocity-time graph is a straight line.
(d) For constant velocity (i.e. acceleration is zero), the graph obtained is a straight line AB parallel to the x-axis (time).
(e) For constant acceleration, the graph obtained is oblique.
(a)Change in velocity for a given time interval is the area enclosed between the acceleration–time graph and time axis.
(b) For constant acceleration, the obtained graph is a straight line parallel to the x-axis, i.e. time (t).
(c) If the acceleration is non-uniform, then the graph is oblique.
Inference: Displacement-time graph for uniformly accelerated or retarded motion is a parabola. Since, for constant acceleration, then the relation between displacement and time is s = ut ± ½ at2 which is quadratic in nature. Thus, displacement–time graph will be parabolic in nature.
(a) The motion of an object is dependent on observation.
(b) A motion is a relative term.
(c) An observation of motion is always with respect to the frame of reference.
Relative Velocity (Introduction to motion in 2D)
Consider two bodies A and B travelling with velocities VAO and VBO, respectively, with respect to origin O, then the relative velocity of B with respect to an observer A, VBA, is given as follows:
VBA = VBO – VAO
Similarly, VAB = VAO – VBO
Thus the relative velocity of any two bodies moving from the same origin is equal to the vector difference of their velocities.
The relative rate of change of VBA gives the relative acceleration of B with respect to A and is given by aBA = aBO – aAO and aAB = aAO – aBA
Fact: Distance between two objects with respect to is independent of the reference frame. If ‘x’ is the minimum distance between the two objects at the time ‘t’ then in any frame of reference the minimum distance of the objects remains constant at the time ‘t’.