Question

A body starts from rest and moving with constant acceleration covers a distance d1 in the 4th second and d2 in the 5th second. What will be the ratio of d1/d2- ?

2/3
4/7
7/9
5/7​

Answer 1

Explanation:

u = 0 m/s

Let acceleration be = a

Displacement in nth second

$= u + \frac{a(2n - 1)}{2}$

$d1 = 0 + \frac{a(2 \times 4 - 1)}{2} = \frac{7a}{2}$

$d2 = 0 + \frac{a(2 \times 5 - 1)}{2} = \frac{9a}{2}$

$\frac{d1}{d2} = \frac{ \frac{7a}{2} }{ \frac{9a}{2} } = \frac{7}{9}$

Therefore, it is your answer.

I hope it helps you. If you have any doubts, then don't hesitate to ask.

Answer 2

Answer:-

7/9

Explanation:-

In this case :-

• As the body is moving with constant

acceleration, it will be same in both

cases.

• Initial velocity of the body will be zero

as it was initially at rest.

We know that distance travelled by a body in the nth second is given by :-

=> Sₙₜₕ = u + a/2(2n - 1)

Where:-

• Sₙₜₕ is distance covered in nth second

• u is initial velocity of the body

• a is acceleration of the body

• n is the given second

For n = 4th sec :-

=> d₁ = u + a/2(2×4-1)

=> d₁ = 0 + a/2(7)

=> d₁ = 7a/2

=> d₁ = 3.5a ----(1)

For n = 5th sec :-

=> d₂ = u + a/2(2×5-1)

=> d₂ = 0 + a/2(9)

=> d₂ = 9a/2

=> d₂ = 4.5a -----(2)

Now, we shall divide eq.1 by eq.2 to get the required ratio :-

=> d₁/d₂ = 3.5a/4.5a

=> d₁/d₂ = 7/9

Thus the ratio of d₁/d₂ will be 7/9 .