TS EAMCET · Physics · Laws of Motion
Amotorcarmovingwithvelocity \(7 \mathrm{~m} / \mathrm{sstopsin} 10 \mathrm{mdistance}\) when brakes are applied. What is the relation between the resistance force \((\mathrm{R})\) and the weight \((\mathrm{W})\) of the car? \(\left(\right.\) Take value of \(g=9.8 \mathrm{~m} / \mathrm{s}^2\) )
- A \(\mathrm{R}=\mathrm{W}\)
- B \(\mathrm{R}=-\mathrm{W}\)
- C \(\mathrm{R}=-\frac{\mathrm{W}}{2}\)
- D \(\mathrm{R}=-\frac{\mathrm{W}}{4}\)
Answer & Solution
Correct Answer
(D) \(\mathrm{R}=-\frac{\mathrm{W}}{4}\)
Step-by-step Solution
Detailed explanation
From equation of motion \[ \begin{aligned} & v^2=u^2+2 a s \\ & \Rightarrow \quad a=\frac{v^2-u^2}{2 s} \\ & =-\frac{(7)^2+(0)^2}{2 \times 10}=-\frac{49}{20}=-2.45 \mathrm{~m} / \mathrm{s}^2 \end{aligned} \] \[ =-\frac{\mathrm{g}}{4} \mathrm{~m} / \mathrm{s}^2 \] We have…
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