IIT-JEE 2011 - Career Point

For (x – a) 2 + (y – b) 2 = r 2 , the motion is in a circular
path with centre at (a, b) and radius r
2
2
x y
2 +
2 = 1 is equation of an ellipse
a b
x × y = constant give a rectangular hyperbola.
Note : To decide the path of motion of a body, a
relationship between x and y is required.
Area under-t graph represents change in velocity.
Calculus method is used for all types of motion (a = 0
or a = constt or a = variable)
s = f(t)
Differentiate w.r.t
time
v = f(t)
integrate w.r.t.
time
Differentiate w.r.t
time
a = f(t)
integrate w.r.t.
time
S stand for displacement
2
dv dv d s
a = v = = ds dt 2
dt
dx dy
Also v x = ⇒ vy = dt dt
2
dv
a x = x d x dv 2 y d y
= and ay = =
dt 2
dt
dt 2
dt
The same concept can be applied for z-co-ordintae.
Projectile motion :
P
ucosθ
u ucosθ
usinθ ucosθ
θ
ucosθ
ucosθ
Q θ
u
usinθ
g
g F
F
g
F
g
F
Projectile motion is a uniformly accelerated motion.
For a projectile motion, the horizontal component of
velocity does not change during the path because
there is no force in the horizontal direction. The
vertical component of velocity goes on decreasing
with time from O to P. At he highest point it becomes
zero. From P to Q again. the vertical component of
velocity increases but in downwards direction.
Therefore the minimum velocity is at the topmost
point and it is u cos θ directed in the horizontal
direction.
The mechanical energy of a projectile remain
constant throughout the path.
the following approach should be adopted for solving
problems in two-dimensional motion :
Resolve the 2-D motion in two 1-D motions in two
mutually perpendicular directions (x and y direction)
Resolve the vector quantitative along these
directions. Now use equations of motion separately
for x-direction and y-directions.
If you do not resolve a 2-D motions in two 1-D
motions in two 1-D motion then use equations of
motion in vector form
v r = u r + at ; s r 1 r
= ut + a t
2
; v r . v r – u r . u = 2 a r s r
2
s = 2
1 ( u
r + v
r )t
When y = f(x) and we are interested to find
(a) The values of x for which y is maximum for
minimum
(b) The maximum/minimum values of y then we may
use the concept of maxima and minima.
Problem solving strategy :
Motion with constant Acceleration :
Step 1: Identify the relevant concepts : In most
straight-line motion problems, you can use the
constant-acceleration equations. Occasionally,
however, you will encounter a situation in which the
acceleration isn't constant. In such a case, you'll need
a different approach
dυ
a x = x d ⎛ dx ⎞ d x
= ⎜ ⎟ =
dt dt
2
⎝ dt ⎠ dt
Step 2: Set up the problem using the following steps:
You must decide at the beginning of a problem
where the origin of coordinates are usually a
matter of convenience. If is often easiest to place
the particle at the origin at time t = 0; then x 0 = 0.
It is always helpful to make a motion diagram
showing these choices and some later positions of
the particle.
Remember that your choice of the positive axis
direction automatically determines the positive
directions for velocity and acceleration. If x is
positive to the right of the origin, the v x and a x are
also positive toward the right.
Restate the problem in words first, and then
translate this description into symbols and
equations. When does the particle arrive at a
certain point (that is, what is the value of t)?
where is the particle when its velocity has a
specified value (that is, what is the value of x
when v x has the specified value)? "where is the
motorcyclist when his velocity is 25m/s?"
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XtraEdge for IIT-JEE 29 MAY 2010