Question

A particle starts from rest moving along a straight line with a constant acceleration of 4 ms−2 . The ratio of the distances travelled in the 2nd and 3rd seconds is


SAY THE RATIO ​

Answer 1

Explanation:

The displacement of the particle in nth second S

n

=u+

2

1

a(2n−1)

where all the symbols have their usual meaning.

Given : u=0

⟹S

n

=

2

1

a(2n−1)

Thus displacement in the 3rd second S

3

=

2

1

a(2×3−1)=

2

5a

Displacement in the 4th second S

4

=

2

1

a(2×4−1)=

2

7a

Percentage increase in the displacement

S

3

ΔS

×100=

S

3

S

4

−S

3

×100

S

3

ΔS

×100=

2

5a

2

7a

−

2

5a

×100=40