Question

a body starts from rest and is accelerated uniformly. find the ratio of the distance travelled by the body in 1st ,3rd and 5th seconds​

Answer 1

Answer:

▪ $\bf{d_1:d_3:d_5=1:5:9}$

Solution:

✴ Given :-

• Initial velocity of body = zero
• Acceleration is constant throughout the whole journey.

✴ To Find :-

• Ratio of distance travelled by the body in first, third and fifth seconds

✴ Formula used :-

▪ Formula of distance covered by body in nth second is given by

$\rightarrow\bf\:d=u+\dfrac{a}{2}(2n-1)$

✴ Terms indication :-

• u denotes initial velocity of body
• a denites acceleration
• d denotes distance
• n denotes no. of second

✴ Calculation :-

→ Distance travelled in first second

$\mapsto\sf\:d_1=0+\dfrac{a}{2}(2-1)\\ \\ \mapsto\bf\:d_1=\dfrac{a}{2}$

→ Distance travelled in third second

$\mapsto\sf\:d_3=0+\dfrac{a}{2}(6-1)\\ \\ \mapsto\bf\:d_3=\dfrac{5a}{2}$

→ Distance travelled in fifth second

$\mapsto\sf\:d_5=0+\dfrac{a}{2}(10-1)\\ \\ \mapsto\bf\:d_5=\dfrac{9a}{2}$

→ Ratio of distance travelled by body in first, third and fifth second is given by

$\underline{\underline{\bf{d_1:d_3:d_5=1:5:9}}}$

Answer 2

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QUESTION :

a body starts from rest and is accelerated uniformly. find the ratio of the distance travelled by the body in 1st ,3rd and 5th seconds.

SOLUTION :

The body starts from rest.

So , U = 0

$D_{n} = u + \dfrac{1}{2} \times ( 2n - 1 )$

Hence,

In this case,

$D_{n} = \dfrac{1}{2} \times ( 2n - 1 )$

Values of n are :

1, 3 and 5

=> $D_{1} = \dfrac{1}{2} \times ( 2 \times 1 - 1 )= \dfrac{1}{2}$

=> $D_{3} = \dfrac{1}{2} \times ( 2 \times 3 - 1 )= \dfrac{5}{2}$

=> $D_{5} = \dfrac{1}{2} \times ( 2 \times 5 - 1 )= \dfrac{9}{2}$

Hence the required ratio becomes :

=> { 1 / 2 } : { 5 / 2 } : { 9 / 2 }

=> 1 : 5 : 9

Answer :

The ratio of the distance travelled by the body in 1st ,3rd and 5th seconds is 1 : 5 : 9.