Q143 / 150 · MHT CET 2025 (26 Apr Shift 2) · Physics

A car moving at a speed ' $V$ ' is stopped in a certain distance when the breaks produce a deceleration 'a'. If the speed of the car is 'nv', what must be the deceleration of the car to stop it in the same distance and in the same time?

  • A
    $\sqrt{\mathrm{n}} \cdot \mathrm{a}$
  • B
    n.a
  • Correct answer
    C
    $\mathrm{n}^2 \cdot \mathrm{a}$
  • D
    $\mathrm{n}^3 \cdot \mathrm{a}$

Solution

A car stops from speed \(v\) under deceleration \(a\). Using \(v=a t \quad \text { and } \quad s=\frac{1}{2} v t\) (the distance-time relations during uniform deceleration). Now the new speed is \(n v\). Let required deceleration be \(\boldsymbol{A}\). For same stopping time: \(n v=A t \quad \Rightarrow \quad A=n a\) For same stopping distance: \(s=\frac{1}{2} v t=\frac{1}{2}(n v) t^{\prime}\) But time is same, so: \(\frac{1}{2} v t=\frac{1}{2} n v t \cdot \frac{a}{A}\) Or use direct formula for distance: \(\frac{v^2}{2 a}=\frac{(n v)^2}{2 A}\) Thus, \(\frac{v^2}{a}=\frac{n^2 v^2}{A} \Rightarrow A=n^2 a .\) \(n^2 a\)

Hence, the correct answer is option: 'C'.

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