A car accelerates from rest with acceleration 1.2 m/s2. A bus moves with constant speed of 12 m/s in a parallel lane. How long does the car take from its start to meet the bus?

To find when the car meets the bus, we equate their displacements from the starting point. The car starts from rest ($u = 0$) with a constant acceleration ($a = 1.2 \text{ m/s}^2$). Using the kinematic equation $s = ut + \frac{1}{2}at^2$, the car's displacement is $s_{car} = 0.5 \times 1.2 \times t^2 = 0.6t^2$ [t3][t5]. The bus moves at a constant speed ($v = 12 \text{ m/s}$), so its displacement is $s_{bus} = vt = 12t$ [c1][c2]. Setting $s_{car} = s_{bus}$ gives $0.6t^2 = 12t$. Dividing both sides by $t$ (since $t > 0$), we get $0.6t = 12$. Solving for $t$ yields $t = \frac{12}{0.6} = 20$ seconds. Thus, the car takes 20 seconds to catch up with the bus. This application of motion equations is a standard method for solving relative motion and intercept problems in kinematics [t3][t6].

Sources

1. [1] Science-Class VII . NCERT(Revised ed 2025) > Chapter 8: Measurement of Time and Motion > 8.3 Speed > p. 113
2. [2] Science-Class VII . NCERT(Revised ed 2025) > Chapter 8: Measurement of Time and Motion > 8.4 Uniform and Non-uniform Linear Motion > p. 117