Equations of Motion: Practice Problem & Solution
A truck moving with velocity \(36\text{ km/hr}\) is stopped by applying brakes in \(2\text{ s}\). If same truck moves with speed \(144\text{ km/hr}\) and brakes are applied then the stopping time will be (Assume the same retardation in both cases)
Solution Explained:
To solve this problem, we apply the core principles of Equations of Motion. Understanding the underlying formula is key to arriving at the correct answer below:
From \(v = u - at\), for stopping \(v = 0\), which gives \(t = u/a\). Since retardation \(a\) is constant, stopping time \(t \propto u\). Since the speed increases by a factor of \(144/36 = 4\), the stopping time becomes \(4 \times 2 = 8\text{ s}\).
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