Question

HI Friends
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​

#Keep smiling :)

Answer 1

Answer:

S1= a(1-1/2) = a/2

S3=a(3-1/2=5a/2

S5=a(5-1/2)=9a/2

so the ratio is 1:5:9

Answer 2

Aɴꜱᴡᴇʀ

$\large \tt{ \orange{d_1:d_3:d_5=1:5:9}}$

_________________

Gɪᴠᴇɴ

• Initial velocity of body = zero

• Acceleration is constant throughout the whole journey.

_________________

ᴛᴏ ꜰɪɴᴅ

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

_________________

Sᴛᴇᴘꜱ

Here we use the formula of distance covered by a body in nth second,that is

$\large\tt {\orange{ \leadsto{d=u+\dfrac{a}{2}(2n-1)}}}$

Where,

• u denotes initial velocity of body

• a denites acceleration

• d denotes distance

• n denotes no. of second

So the distance travelled in first second is,

$\begin{lgathered}\leadsto\tt\:d_1=0+\dfrac{a}{2}(2-1)\\ \\ \dashrightarrow\sf{ \red{d_1=\dfrac{a}{2}}}\end{lgathered}$

And the distannce travelled in third second is,

$\begin{lgathered} \leadsto\tt\:d_3=0+\dfrac{a}{2}(6-1)\\ \\ \dashrightarrow\sf{ \green{d_3=\dfrac{5a}{2}}}\end{lgathered}$

Finally the distance travelled in fifth second

$\begin{lgathered}\leadsto\sf\:d_5=0+\dfrac{a}{2}(10-1)\\ \\ \dashrightarrow\orange{d_5=\dfrac{9a}{2}}\end{lgathered}$

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

$\large{\pink{ \mapsto\sf{d_1:d_3:d_5=1:5:9}}}$

_________________

$\huge{\mathfrak{\purple{hope\; it \;helps}}}$