COMEDK · Physics · 3. Motion In One Dimension
A car, starting from rest, accelerates at the rate (f) through a distance (S), then continues at constant speed for some time \((\mathrm{t})\) and then decelerates at the rate \(\dfrac{f}{2}\) to come to rest. If the total distance is 5 S , then
- A \(S=\dfrac{1}{2} f t^2\)
- B \(S=4 f t^2\)
- C \(S=\dfrac{1}{4} f t^2\)
- D \(S=2 f t^2\)
Answer & Solution
Correct Answer
(A) \(S=\dfrac{1}{2} f t^2\)
Step-by-step Solution
Detailed explanation
Let the car accelerate from rest at rate \(f\) for distance \(S\). The final velocity \(v\) reached is given by \(v^2 = 0^2 + 2fS\), so \(v = \sqrt{2fS}\). The time taken for this phase is \(t_1 = \dfrac{v}{f} = \dfrac{\sqrt{2fS}}{f} = \sqrt{\dfrac{2S}{f}}\). The car then…
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