Question

Abody travels for 15 sec starting from rest with
constant acceleration. If it travels distances S,,S,
and S, in the first five seconds, second five sec-
onds and next five seconds respectively the re-
lation between S, S, and-s, is:
(1) S = S₂ =S; (2) 55 = 35₂ = S₃
23) ss =
Then​

Answer 1

Answer:

$15s_{1}$ = $5s_{2}$ = $3s_{3}$

Explanation:

First lets see what is the data we have

• time is 15s

• acceleration is constant. so lets assume its "a"

• distance in first 5 seconds is $s_{1}$

• distance in second 5 seconds is $s_{2}$

• distance in third 5 seconds is $s_{3}$
• body started to move from rest

so lets start solving the question:

1. For first five seconds :

t = 5

s = $s_{1}$

u = 0

a = a

$s_{1}$ = $ut + 0.5at^{2}$

put the values we get

$s_{1}$ = $\frac{25a}{2}$

2. For the second 5 seconds :

distance traveled in 10 seconds ⇒

t = 10

s = $s_{10}$

u = 0

a = a

$s_{10}$ = $ut + 0.5at^{2}$

put the values we get

$s_{10}$ = $\frac{100a}{2}$ = 50a

distance traveled in second 5 seconds = distance traveled in 10 seconds - distance traveled in first 5 seconds

$s_{2}$ = $s_{10}$ - $s_{1}$

                       = 50a - (25/2)a

                       = $\frac{75a}{2}$

3. For the Third 5 seconds :

distance traveled in 15 seconds ⇒

t = 15

s = $s_{15}$

u = 0

a = a

$s_{15}$ = $ut + 0.5at^{2}$

put the values we get

$s_{15}$ = $\frac{225a}{2}$

distance traveled in third 5 seconds = distance traveled in 15 seconds - distance traveled in first 10 seconds

$s_{3}$ = $s_{15}$ - $s_{10}$

                       = $\frac{225a}{2} - 50a$

                       = $\frac{125a}{2}$

To find relation :

$s_{1}$ = $\frac{25a}{2}$

$s_{2}$ = $\frac{75a}{2}$ = $3s_{1}$

$s_{3}$ = $\frac{125a}{2}$ = $5s_{1}$

$s_{1}$ : $s_{2}$ : $s_{3}$ = 1:3:5

and

$15s_{1}$ = $5s_{2}$ = $3s_{3}$

Answer 2

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