Relative Motion in One Dimension: Practice Problem & Solution
A bus starts from rest moving with an acceleration of \(2\text{ m/s}^2\). A cyclist, 96 m behind the bus starts simultaneously towards the bus at a speed of \(20\text{ m/s}\). After what time will bus overtake the cycle :
Solution Explained:
To solve this problem, we apply the core principles of Relative Motion in One Dimension. Understanding the underlying formula is key to arriving at the correct answer below:
Let the bus's position be \(x_b = (1/2)at^2 = t^2\). The cyclist's position (starting 96m behind) is \(x_c = 20t - 96\). For meeting, \(t^2 = 20t - 96 \Rightarrow t^2 - 20t + 96 = 0\). Factoring gives \((t-8)(t-12) = 0\), so \(t=8\)s or \(t=12\)s. At \(t=8\)s, cyclist is faster than bus. At \(t=12\)s, bus is faster, so it overtakes.
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