Question

A body starting from rest moves with constant acceleration, then the ratio of the distance covered in 5th second to 5 seconds is?

Answer 1

Distance covered in nth second
S = u + a(n - 0.5)
In 5th second
S₁ = 0 + a(5 - 0.5)
S₁ = 4.5a


Distance covered in t seconds
S = ut + 0.5at²
In 5 seconds
S₂ = 0 + (0.5 × a × (5 s)²)
S₂ = 12.5a

S₁ / S₂ = 4.5a / (12.5a)
= 9 / 25

Ratio of distance travelled in 5th second to 5 seconds is 9 : 25

Answer 2

Given,
u = initial velocity =0

acceleration = a

Sn = displacement in the n th second.

S5 = u + a ( n -1/2)

S 5 = 0 + a (5-1/2)

S 5 = a (9/2)

Distance covered in 5 seconds

s=ut+1/2at²

=0+1/2*5²a

=25/2a

Ratio of distance covered in 5th second to 5 seconds =9/2a:25/2a

=9:25.