Question

Pls answer it with process...​

Answer 1

Solution:-

Given:-

For Acceleration;

u = 0 ( It starts from rest)

t = 5 second.

Acceleration = a.

By The Formula,

v = u + at

=> v = 0 + a*5

=> v = 5a.

Again, For Retardation;

u = 5a.

v = 0 ( It comes to rest).

Acceleration = -a

Now,

Average Speed = Total Distance/ Total Time Taken

By the Formula of Distance,

d = ut + 1/2 at²

=> A. S = [ ( ut + 1/2 at²) + ( ut - 1/2 at²)] / 10 sec.

=> A.S. = [ ( 0*5 + 1/2*a*5²) + ( 5a*5 - 1/2*a*5²)] / 10

=> A.S = [ 1/2*a*25 + 25a - 1/2*a*25)] / 10

Taking L.C.M.

=> A. S. = [ (25a + 50a - 25a)/2]/10

=> A.S. = 50a / 10*2

=> A.S. = 50a / 20

=> A.S = 2.5a.

Now,

Ratio of Average Speed and the Maximum speed :-

Average Speed =2.5a

Maximum Speed = 5a.

Hence,

Ratio = 2.5a/5a

=> Ratio = 1/2

=> Ratio = 1:2.

Hence,

Option "A" is Correct!!

Answer 2

$\huge{\mathfrak{Solution:-}}$

$\sf{In\;the\;first\;case\;where\;object\;is\;accelerating,}$

$\bf{Given:-}$

$\sf{Intial\;velocity = 0\;m/s}$

$\sf{Time\;taken = 5\;sec.}$

$\sf{Let\;acceleration\;be\;a\;then,}$

$\sf{v = u + at}$

$\sf{v = 0 + a\times 5}$

$\sf{v = 5a}$

$\sf{In\;second\;case\;where\;object\;is\;decelerating,}$

$\bf{Given:-}$

$\sf{Intial\;velocity\;u' = 5a}$

$\sf{Final\;velocity\;v' = 0\;m/s}$

$\sf{Time\;taken = 10\;sec}$

$\sf{Acceleration=-a}$

$\sf{Average\;speed = \frac{Total\;distance}{Total\;time}}$

$\sf{Total\;time = 10\;sec}$

$\sf{So,\;first\;we\;find\;total\;distance,D}$

$\sf{D = Acceleration+Deceleration}$

$\sf{D = ut+1/2at^{2}+u't'+1/2a't'^{2}}$

$\sf{D = 0+1/2\times a\times 25+5a\times 5+1/2\times (-a)\times 100}$

$\sf{D = 25a}$

$\sf{Average\;speed = \frac{Total\;distance}{Total\;time}}$

$\sf{Average\;speed=\frac{25a}{10}=2.5a}$

$\sf{So,\;the\;ratio\;between\;average\;speed\;and\;maximum\;speed\;is,}$

$\sf{\implies{2.5a:5a}}$

$\sf{\implies 1:2}}$

$\boxed{\sf{So,\;ratio\;between\;Average\;speed\;and\;max.\;speed = 1:2}}$