Question

A particle starts from rest find the distance
travelled by it in 5th half of the second if its
- acceleration is 2 m/s
(1) 4m
(2) 6.25 m
(3) 225 m
(4) Zero​

Answer 1

Explanation:

By this the actually mean when the particle is moving the distance traveled in 5th half second i.e only distance travelled in the half of the 5th second

For this the formula is S(nth)=u+a/2 (2n-1) here n is 5 seconds

Therefore the answer we get is 2.25 m by substituting values in it respectively

Answer 2

$\huge\mathfrak\pink{\underline{\underline{AnSwer}}}$

★ We can solve it by applying LAW OF ODD NUMBERS:-

→ If a body starts from rest with a constant acceleration then the ratio of successive distances covered in equal interval of time will be the ratio of odd numbers.

Here,

S1st : S2nd : S3rd

1. : 3. : 5

Now, distance covered by body in 5th 1/2 sec will be

S5th half / S1st half. = ( 2×5-1)/(2×1-1)

S5th half / S1st half = 9/ 1

=> S5th half = 9× S1st half

=> S5th half = 9× (ut + 1/2 a t^2)

=> S5t half = 9× (0+ 1/2 *2*1/2 *1/2)

=> S5th half = 9× 1/4

=> Distance travelled by 5th half is 9/4m

● Your answer is option 3 ...i.e 2.25m●

Happy to help!

:)