TS EAMCET · Physics · Motion In One Dimension

A man of mass 80 kg goes to the market on a scooter of mass 100 kg with certain speed. On application of brakes, the stopping distance is \(S_1\). The man returns home on the same scooter, with the same speed with a 60 kg bag of rice. If \(S_2\) is the new stopping distance when the brakes are applied with the same force, then

A \(7 S_1=4 S_2\)
B \(2 S_1=S_2\)
C \(3 S_1=4 S_2\)
D \(4 S_1=3 S_2\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(4 S_1=3 S_2\)

Step-by-step Solution

Detailed explanation

When stopping distance is \(\mathrm{s}_1\) \(v^2-u^2=2 a_1 s_1 \Rightarrow 0-u^2=2 \frac{F}{100+80} s_2 \qquad ...\mathrm{(i)}\) When stopping distance is \(\mathrm{s}_2\) \(v^2-u^2=2 a_2 s_2 \Rightarrow 0-u^2=2 \frac{F}{100+80+60} \times s_2 \qquad ...\mathrm{(ii)}\) From eq.…

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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