Question

'A particle starts from rest under constant
acceleration a = 4 m/s2. Ratio of respective
distance travelled by the particle in 3rd and 5th
second of its motion is
09:25
(12) 5:9
(3) 9:5
(4) 25:9​

Answer 1

(2) is the correct answer.

Given:

u = 0 since particle starts at rest

a = 4 m/s² (constant)

To find:

distance travelled in 3rd second: distance travelled in 5th second

Solution:

We know that

s = ut + (1/2)at²

Now,

since the particle starts from rest, u = 0

⇒ For our purposes, s = (1/2)at²

Now,

distance travelled in 3rd second = distance till 3 seconds - distance till 2 seconds

⇒ distance in 3rd second = (1/2)(4)(3²) - (1/2)(4)(2²) = 18 - 8 = 10 m

Also,

distance travelled in 5th second = distance till 5 seconds - distance till 4 seconds

⇒ distance in 5th second = (1/2)(4)(5²) - (1/2)(4)(4²) = 50 - 32 = 18 m

⇒ ratio of distances = 10:18 = 5:9

⇒ ratio is 5:9

∴ (2) is the correct answer.

SPJ2

Answer 2

Answer:

The right response is (2) 5:9.

Explanation:

Given:

Since the particle is at rest, u equals 0.

a = 4 m/s² (constant)

In search of:

the distance covered in the third second; the distance covered in the fifth second

Solution:

Knowing this

s = ut + (1/2)at²

Now,

u = 0 because the particle starts at rest.

S equals (1/2)at2 for our needs.

Now,

Distance travelled in a third second is equal to the difference between the previous two and third seconds.

Distance in 3 seconds equals (1/2)(4)(32) - (1/2)(4)(22) = 18 - 8 = 10 m

Also,

Distance travelled in the fifth second is equal to the difference between the distances travelled in the previous four seconds and the fifth second.

Distance in 5th second = (1/2)(4)(52)-(1/2)(4)(42) = 50 - 32 = 18 m

Distances are proportionately 10:18 to 5:9

ratio is 5:9

#SPJ5