Question

A freely falling body describes a distance of x in the first 2 seconds and a distance of y in next 2 seconds, then relation between x and y is​

Answer 1

Answer:

• Relation between x and y will be y = 3 x

Explanation:

First case : When freely falling body describes a distance of x in first 2 sec

• Initial velocity of body = 0
• distance covered = x
• time taken = 2 s
• acceleration due to gravity = 9.8 m/s²

Using Second equation of motion

→ s = u t + 1/2 a t²

→ x = ( 0 ) ( 2 ) + 1/2 ( 9.8 ) ( 2 )²

→ x = 19.6 m

Now,

Calculating final velocity of body after covering a distance x

Using first equation of motion

→ v = u + a t

→ v = ( 0 ) + ( 9.8 ) ( 2 )

→ v = 19.6 m/s

hence, final velocity of body after covering a distance x will be 19.6 m/s.

Second case : When the same object covers a distance y in next 2 sec

• Initial velocity in this case = 19.6 m/s
• distance covered = y
• time taken = 2 s
• acceleration due to gravity = 9.8 m/s²

Using second equation of motion

→ s = u t + 1/2 a t²

→ y = ( 19.6 ) ( 2 ) + 1/2 ( 9.8 ) ( 2 )²

→ y = 58.8 m

so,

Calculating the relation between x and y

→ x / y = 19.6 / 58.8

→ x / y = 1 /3

→ y = 3 x

therefore,

• Relation between x and y will be y = 3 x

Answer 2

$\huge \bf \red{ \mid{ \overline{ \underline{ANSWER}}} \mid}$

As in the question given , we have to find the realation between 'x' and 'y' in two events

Hence , we will have two cases!

$\bf{Let's \: solve \: the \: First \: case }$

$\bf \pink{GIVEN - }$

→ initial velocity of the body in respect to the observer ( let's say u ) - 0.

→ distance - 'x'

→ time = 2 sec

→ acceleration ( here the acceleration will be the acceleration due to gravity ) = 9.8 m/s

→ x = ut + 1/2 at²

Here

x = distance

u = intitial velocity

a = acceleration

t = time

→ x = (0) × 2 + 1/2 a(2)

→ x = 1/2 × 9.8 (2)²

→ x = 1/2 × 9.8 × 4

→ x = 19.6 m

We found the value of x for first two seconds ,

let's find it's final velocity

→ V = u + at

Here , v is final velocity

→ v = 0 + 9.8 (2)

→ v = 19.6 m/s

$\bf{ \: second \: case}$

what We got from first case are -

→ v = 19.6 m/s

→ x = 19.6 m

now ,

initial velocity for the second case ( say u' ) will be the final velocity of first case

Hence , u' = 19.6 m/s

t' = 2 sec

a = 9.8 m/s

→ y = ut + 1/2 at²

→ y = 19.6 (2) + 1/2 × 9.8 (2)²

→ y = 39.2 + 19.6

→ y = 58.8 m

Hence ,

The object covered 39.2 metres more in next 2 seconds !

Now , the relation

→ y = 3(19.6)

→ y = 3x