FRAME OF REFERENCE: The fixed point or place with respect to which the position, velocity, acceleration etc. of another body or an object is measured is known as frame of reference.
CARTESIAN COORDINATE SYSTEM:
1. One dimensional reference frame: A particle P is moving along X axis. The distance of it from point O is x. So, point O is the point of reference of particle P.
2. Two dimensional reference frame: If a particle P is placed on a plane, then the position of it is defined by two mutual perpendicular axes OX and OY respectively. Let us consider the coordinates of point P are (x, y). Then OX and OY axes represent two dimensional reference frame of particle P with O as origin.
Let PA and PB are the perpendiculars on OX and OY axes respectively.
So, OA = x, OB = PA = y and OP = r (say).
r = OP = 
= 
.
If 
is the angle of OP with OA then 
.
3. Three dimensional reference frame: If a particle P (example – flying insect) is placed in three dimensional space then the position of it is represented by three dimensional reference frame.
Let us consider OX, OY and OZ are three perpendicular axes, where O is the origin.
Let the coordinates of point P are (x, y, z) and 
is the position vector of P.
r = OP = 
= 
.
Cartesian coordinate system is of two types.
(i) Inertial frame of reference: Any frame of reference which is either at rest or moving with uniform velocity is called inertial frame of reference. In inertial frame of reference Newton’s first law of motion is valid.
(ii) Non- inertial frame of reference: A frame of reference which is accelerated is called non- inertial frame of reference. In non-inertial frame of reference Newton’s first law of motion is not valid.
- CYLINDRICAL COORDINATE SYSTEM: Let us consider OX, OY and OZ are three mutually perpendicular axes, where O is the origin. Let the particle is at point P with coordinates (
, 
, z). Then x = cos , y = cos and z = z where x, y and z are the Cartesian coordinates of point P.
r = 
and = 
.
- SPHERICAL COORDINATE SYSTEM: It is the three dimensional frame of reference where the point P is represented by coordinates (r,
, 
). Where OP = r, is the angle of OP with Z axis and is the angle between projection of OP on XY plane and OX axis.
Then x = r sin cos , y = r sin sin and z = r cos .
Instantaneous speed: The instantaneous speed at a point is defined as the limit of the average speed over a path length that tends to become zero but always includes the point.
Let us consider a particle moves along a path ABCD and we want to calculate the instantaneous speed of it at point C. As the path 
s is very small, the corresponding time interval t also small.
The instantaneous speed v is thus given by v = = 
.
Example: A juggler juggles balls throwing upwards. He throws one ball when the previous one is at a height halfway of its journey. How high do the balls rise if he throws n balls per second?
Let us consider juggler juggles balls throwing upwards with speed u and maximum height travelled by the ball is h.
Time interval between two balls is 
s. so, 
= u() – 
g()2 or, = 
– 
or, u = 
+ 
At maximum height v = 0 using relation v2 = u2 – 2as we get, h = 
= 
.
Or, 2gh = ( + )2
Or, 8gh = h2n2 + 2gh + 
Or, 6gh = h2n2 +
Or, h2n4 – 6ghn2 + g2 = 0
h = 
= 
= 
.
- A particle starting from origin O(0,0,0) moves to point P (0,0,a) then moves point Q (0,a,a) then point R (0,a,0) then point S (a,a,0) and finally point T (a, a, a). Find the distance and displacement of the particle.
If the origin is O then distance s = OP + PQ + QR + RS + ST = a + a + a + a + a = 5a.
Displacement is OT = 
= 
= 
a.
- Two balls A and B are projected vertically upwards with speed u and v from the top and the bottom of a tower of height h respectively. Find the meeting time of the balls [Neglect air resistance]
(a) Above the top of the tower:
Let us consider the balls meet after time t and taking upward motion is positive.
For ball A displacement = s1, initial speed = u and acceleration = -g.
Then, S1 = ut – gt2 —– (i)
For ball B displacement = s2, initial speed = v and acceleration = -g.
Then, S2 = vt – gt2 ——(ii)
From equation (i) and (ii) we get, S1 + h = S2
Or, ut – gt2 + h = vt – gt2 t = 
.
(b) In between top and bottom of the tower:
Let us consider the balls meet after time t and taking upward motion is positive.
For ball A displacement = – S1, initial speed = u and acceleration = -g.
Then, – S1 = ut – gt2 or, S1 = – ut + gt2 —– (i)
For ball B displacement = s2, initial speed = v and acceleration = -g.
Then, S2 = vt – gt2 ——(ii)
From equation (i) and (ii) we get, S1 + S2 = h
Or, – ut + gt2 + vt – gt2 = h
t = .
(c) Below the bottom of the tower:
Let us consider the balls meet after time t and taking upward motion is positive.
For ball A displacement = – S1, initial speed = u and acceleration = -g.
Then, – S1 = ut – gt2 or, S1 = – ut + gt2 —– (i)
For ball B displacement = – s2, initial speed = v and acceleration = -g.
Then, – S2 = vt – gt2 or, S2 = – vt + gt2 ——(ii)
From equation (i) and (ii) we get, S1 – S2 = h
Or, – ut + gt2 + vt – gt2 = h
t = .
CONDITIONS OF MINIMUM DISTANCE BETWEEN TWO BODIES: Two cars A and B are moving along a straight line in same direction separated by a distance. Car B is behind car A. The instantaneous speed of cars A and B are respectively v1 and v2.
If v1 > v2 , then the distance between the cars increases.
If v1 < v2 , then the distance between the cars decreases.
If v1 = v2 , then the distance between the cars is minimum.
EXAMPLE: Two cars A and B (B is behind A) separated by a distance 200m moving along the same direction with initial speed 3ms-1 and 9ms-1 respectively. The acceleration of car A and car B are respectively 4ms-2 and 2ms-2 . When the separation between the cars is minimum and what is the minimum distance?
The distance between the cars is minimum when vA = vB and the time is t.
vA = 3 + 4t and vB = 9 + 2t so, 3 + 4t = 9 + 2t or, t = 3s.
Distance travelled by car A for 3s is sA = 3 
3 + 4 32 = 27 m
Distance travelled by car B for 3s is sB = 9 3 + 2 32 = 36 m
Minimum distance = 200 + sA – sB = 200 + 27 – 36 = 191 m.
.
TWO DIMENSIONAL MOTION:
- A body of mass m is moving with constant speed u along x axis and F force is applied on it along y axis. Find velocity of it after t second and trajectory.
Along x axis: initial speed ux = u acceleration a = 0 velocity after t second
vx = ux = u ——–(i)
and distance travelled x = ut ——-(ii)
Along y axis: initial speed uy = u cos 900 = 0 acceleration a = 
velocity after t second
vy = uy + at = 
——(iii)
and distance travelled y = uyt + at2 = 
——-(iv)
Velocity of the body is v = 
= 
if the velocity of the body creates angle with positive x axis then, tan = 
= 
.
Trajectory: From equation (ii) t = 
and using the value of t in equation (iv) we get, y = 
.
- The initial velocity of a body is and acceleration is . Find the displacement of the body.
Displacement of the body is = t + = ( = (at + c
t2 )
+ (bt + dt2 .
CONCEPT ON MOVING FRAME: When a reference frame is moving and a body is dropped from that frame, then that body also gains the instantaneous speed of that frame.
- A lift moving upwards with speed u. When the lift is at a height h from ground a ball is dropped from it. Find the time taken by the ball to touch the ground.
Let the time is t and taking downward direction is positive. h = – ut + g t2
Or, gt2 – 2ut –2h = 0
Or, t = 
t = 
[as time cannot be negative].
- A lift moving upwards with speed u. When the lift is at a height h from ground a boy throws a ball vertically upwards with speed v from it. Find the time taken by the ball to touch the ground.
Let the time is t and taking downward direction is positive. The speed of the ball from the lift is w = (u + v) then,
h = – wt + gt2
Or, gt2 – 2wt –2h = 0
Or, t = 
t = 
[as time cannot be negative].
- A lift moving upwards with speed u. When the lift is at a height h from ground a boy throws a ball vertically downwards with speed v (v > u) from it. Find the time taken by the ball to touch the ground.
Let the time is t and taking downward direction is positive. The speed of the ball from the lift is w = (v-u) then,
h = wt + gt2 or, gt2 + 2wt –2h = 0 or, t = 
t = 
[as time cannot be negative].
CONCEPT ON VELOCITY:
- The length of the minute hand of a clock is r cm. Calculate the magnitude of change in velocity of the tip of the minute hand from 12 noon to 12.15 pm.
The speed of the minute hand of a clock is v = = = cm
When the minute hand of a clock is at 12noon the direction of velocity ( is towards +ve x axis. When the minute hand of a clock is at 12.15pm the direction of velocity ( is towards -ve y axis. Using vector subtraction method the change in velocity is ( – )
= = = cm. [Magnitude of or is || = ||= ].
- In the figure, the blocks A and B are of equal mass connected by a massless inextensible string over a pulley P which is frictionless. At any instant Block A moves on a horizontal surface with speed VA and string creates angle with it. At that time B is moving with speed VB. Find the relation between VA and VB.
Let the lengths of the sections of the string be AP = l1 and PB = l2 then the total length of the string is (l1 + l2).
In a further time t, let A moves to the right (point A/) by x and B move down (point B/) by y.
A/C is perpendicular on AP.
Now A/P PC = PA – AC = PA – AA/ = l1 – x and PB/ = l2 + y.
So, the total length of the string is A/P + PB/ = l1 – x + l2 + y = (l1 + l2 – x + y).
Therefore, l1 + l2 = l1 + l2 – x + y
Or, x = y
Differentiating both side of equation with respect to time t we get,
Or, =
VA = VB [The velocities of A and B are respectively VA = and VB = .]
- A particle slides without friction from the highest point A of a vertical circle along a chord. If the particle starts from rest, show that the time of descent is independent of the chord chosen.
Let the radius of the circle is R. If the particle slides along the chord AB as shown in the figure, then the component of acceleration due to gravity along AB is g.
The length AB = 2AC = 2AO = 2R [where is the angle made by AB with the vertical diameter.]
Using the equation s = a we get, 2R = g or, t = .
So, the time of descent of the particle is independent of the chord chosen.