Mathematics Definition And Proof Based Problems

### Definition And Proof Based Problems

Q 3253191944

Write the expression for distance covered in n th second by a uniformly accelerated body.

Solution:

If a is the uniform acceleration, then

s = u + a/2 (2n - 1 ) ,

where u is the initial velocity.
Q 3244801753

Explain clearly, with examples, the distinction between :
(a) Magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval ;
(b) Magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For the sake of convenience, consider one-dimensional motion only].

Solution:

(a) Suppose a body moves from point P to point Q along a straight line and returns back to the initial point P along the same path. Here the distance covered by the body = AB + BA = 2AB but the displacement of the body is equal to zero because displacement is given by a vector drawn from initial position to final position of the body.
(b) Suppose the body in above example takes time 't' to complete the whole journey. Then magnitude of average velocity

 = text( magnitude of displacement)/text (time) = 0/t = 0

However average speed =  (2 AB)/t The maximum magnitude of displacement is equal to the distance covered which is the case when the body moves on a straight line without reversing its direction of motion. In all other situations, the magnitude of displacement (hence average velocity) is less than the distance (hence average speed) covered. In no case, the magnitude of displacement (or average velocity) can be more than the distance covered (or average speed).
Q 3264101955

Derive the three equations of motion by calculus method. Express conditions under which they can be used.

Solution:

Consider an object moving in straight line with uniform acceleration = a.
Let at t = 0 velocity of the body = u
at t = t velocity of the body = v
(i) Velocity-time relation: Let dv be the change in velocity in time interval. dt. Then acceleration

 a = (dv)/( dt) or  dv = a \ \ dt

Integrating from 0 -> t when velocity
changes from u -> v

 int_u^v dv = a int_0^t dt

or  v - u = at

or  v = u + at ................(i)

(ii) Distance-time relation : Consider an object moving in a straight line with uniform acceleration 'a'. Let at any instant t, dx be the displacement of the object in time interval at., Then instantaneous velocity v is given by
 v = (dx)/(dt)  or  dx = v dt

or  dx = (u + a t) dt
[from (i) v = u + at ]

Let x_0 = displacement at t = 0
x = displacement at t = t
Integrating within limits

 int_(x_0)^x dx = int_0^t (u + at) dt = u int_0^t dt + a int_0^t t d t

 x - x_0 = ut + 1/2 at^2

or  x = x_0 + ut + 1/2 at^2 ............(ii)

If x - x_0 = s = distance covered by an object in time t then

 s = ut + 1/2 at^2

(iii) Velocity-displacement relation. Consider a particle moving in a straight line with initial velocity u, and uniform acceleration 'a'.

then  a = (dv)/(dt) = (dv)/(dx) xx (dx)/(dt) = v (dv)/(dx)

 adx= vdv

Let u be the velocity of object at position x_o
v be the velocity of object at position x Integrating above within limits

 int_(x_o)^x a dx = int_u^v v dv

 a (x - x_0 ) = v^2/2 - u^2/2

v^2 - u^2 = 2a (x - x_0 )

Putting x - x_0 = s we get

 v^2 - u^2 = 2 as ..........(iii)

The above three laws are valid under the conditions, only when the acceleration is uniform.
Q 3214101959

Derive an equation for the distance covered by a uniformly accelerated body in n th second of its motion. A body travels half its total path in the last second of its fall from rest. Calculate the time of its fall.

Solution:

For a body having a uniform acceleration 'a' in a straight line, starting with an initial velocity u, the displacement in 'n' seconds
is given by,
S_n = n u + 1/2 a n^2

In (n -1) seconds,

 S_(n - 1) = (n -1) u + 1/2 a (n - 1)^2

:.  Displacement in nth second = S_n - S_(n - 1)

 = u + a/2 (2n - 1)

Let S be the complete length of fall and t be the time taken for it. Then,

 S = 1/2 g t^2 ..............(i)

Also ,S/2 is covered in the last second.

:. S/2 = 0 + g/2 ( 2t - 1) ................(ii)

Using (i) and (ii), solve for t to be,

 S = g (2t - 1) = - 1/2 g t^2 ,

i.e., 4 t g - 2 g = g t^2

g t^2 - 4 t g + 2 g = 0

=> t^2 - 4t + 2 = 0

i.e,  t = (4 pm sqrt ( 16 - 8))/2 = (4 pm 2 sqrt2 )/2

 t = 2 pm sqrt2
Q 3284101957

Draw velocity-time graph of uniformly accelerated motion in one dimension. From the velocity time graph of uniform accelerated motion, deduce the equations of motion in distance and time.

Solution:

Consider an object moving along a straight line with uniform acceleration a. Let u be the initial velocity at t = 0 and v be the final velocity after time t.

From graph O A = E D = u
O C = E B = v
O E = t =AD

(i)  s = ut + 1/2 a t^2
area under velocity time graph for a given time interval represents the distance covered by a uniformly accelerated object in a given time interval.
From graph, acceleration, a = slope of velocity-time graph AB.

 :. a = (BD)/(AD) = (DB)/t

or  DB = at
Distance travelled by object in time t is

s = area of trapezium O A B E
= area of rectangle O A D E + Area of triangle A D B

 = O A xx O E + 1/2 D B xx A D

 = u t + 1/2 a t xx t = u t + 1/2 a t^2

(ii) v^2 - u^2 = 2 as
Distance travelled by an object in time interval t is
s = area of trapezium O A B E

 = 1/2 ( E B + O A) xx O E

 = 1/2 ( E B + E D) xx O E

Acceleration,
a = slope of velocity time graph AB

 a = (DB)/(AD) = (EB - ED)/(O E)

 O E = (E B - E D)/a

 s = 1/2 ( E B + E D) xx ( E B + E D) /a

 = 1/(2a) ( E B^2 + E D^2)

 = 1/(2a) ( v^2 - u^2)

 v^2 - u^2 = 2 a s