Solution:

If a is the uniform acceleration, then

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

where `u` is the initial velocity.

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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).

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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.

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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`

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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`

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