A car is moving with constant acceleration after starting from rest. If it travels a distance x in the first 10 sec and the distance y in the next 10, then the ratio of x and y is:

CONCEPT:

Equation of Kinematics:

• These are the various relations between U, V, a, t, and s for the particle moving with uniform acceleration where the notations are used as:
• Equations of motion can be written as

⇒ V = U + at

\(⇒ s =Ut+\frac{1}{2}{at^{2}}\)

⇒ V2 = U2+ 2as

where, U = Initial velocity, V = Final velocity, a = Acceleration, t = time, and h = Distance covered

CALCULATION:

Given U = 0 m/sec

Let the acceleration be a.

By the second equation of motion, the distance covered in the first 10 sec is given as (s₁₀ = x),

\(⇒ s_{10} =0\times10+\frac{1}{2}{\times a\times10^{2}}\)

⇒ x = 50a     -----(1)

The distance covered in the next 10 sec is given as (s2 = y),

⇒ y = Distance covered in first 20 sec - Distance covered in first 10 sec

⇒ y = s₂₀ - s₁₀     -----(2)

By the second equation of motion, the distance covered in the first 20 sec is given as,

\(⇒ s_{20} =0\times10+\frac{1}{2}{\times a\times20^{2}}\)

⇒ s20 = 200a     -----(3)

By equation 1, equation 2, and equation 3,

⇒ y = 200a - 50a

⇒ y = 150a     -----(4)

By equation 1 and equation 4,

\(\Rightarrow \frac{x}{y}=\frac{50a}{150a}\)

\(\Rightarrow \frac{x}{y}=\frac{1}{3}\)

• Hence, option 2 is correct.