Question

The velocity of car A and car B are equal when car A is 30 m behind car B and both the cars are moving in the same direction. The acceleration of car A and car B is 30 m/s² and 20 m/s² respectively. Then after how much time both the cars will meet at the same point.
1. \(\sqrt3\,sec\)
2. \(\sqrt6\,sec\)
3. \(\sqrt2\,sec\)
4. \(\sqrt5\,sec\)

Answer 1

Correct Answer - Option 2 : \(\sqrt6\,sec\)

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 = U_A = U_B, a_A = 30 m/s², and a_B = 15 m/s²

Let after time t both the cars will meet at the same point.

• So distance covered by car A will be,

⇒ s_A = 30 + x     -----(1)

\(⇒ s_{A} =Ut+\frac{1}{2}{a_{A}t^{2}}\)

\(⇒ 30+x =Ut+\frac{1}{2}\times{30\times t^{2}}\)

⇒ 30 + x = Ut + 15t²     -----(2)

• The distance covered by car B is given as,

\(⇒ s_{B} =Ut+\frac{1}{2}{a_{B}t^{2}}\)

\(⇒ x =Ut+\frac{1}{2}\times{20\times t^{2}}\)

⇒ x = Ut + 10t2     -----(3)

By equation 2 and equation 3,

⇒ 30 + Ut + 10t2 = Ut + 15t2

⇒ 5t² = 30

⇒ \(t=\sqrt6\) sec

• Hence, option 2 is correct.