TS EAMCET · Physics · Motion In Two Dimensions

Two cars $A$ and $B$ initially at rest are moving in the same direction with accelerations, $a_{1}$ and $a_{2}$ respectively. After a certain time, they achieve velocities $v_{1}$ and $v_{2}$ respectively and are separated by a distance of $50 m .$ If $(a_{1} - a_{2}) = 4 m s^{- 2},$ then the quantity $(v_{1} - v_{2})$ will be

A $24 m s^{- 1}$
B $20 m s^{- 1}$
C $40 m s^{- 1}$
D $12 m s^{- 1}$

Verified Solution

Answer & Solution

Correct Answer

(B) $20 m s^{- 1}$

Step-by-step Solution

Detailed explanation

Use the first equation of motion, v=u+at, both the cars are initially at rest it means the equations for two different cars moving with difference acceleration after same time, v1=0+a1t & v2=0+a2t⇒v1-v2=a1-a2t ...1 , now the use the second equation…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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Two cars A and B initially at rest are moving in the same direction with accelerations, a1 and a2 respectively. After a certain time, they achieve velocities v1 and v2 respectively and are separated by a distance of 50 m. If a1-a2=4 m s-2, then the quantity v1-v2 will be

Step-by-step Solution