bus starts moving with acceleration 2 m/s2. A cyclist 96 m behind the bus starts simultaneously towards the bus at 20 m/s. After what time will he be able to overtake the bus?4sec8sec18sec16sec

To solve this problem, we need to set up an equation that represents the relative positions of the bus and the cyclist over time.

The bus starts from rest and accelerates at 2 m/s^2, so its position at any time t can be given by the equation:

d_bus = 1/2 * a * t^2, where a is the acceleration of the bus.

The cyclist starts 96 m behind the bus and travels at a constant speed of 20 m/s, so his position at any time t can be given by the equation:

d_cyclist = v * t + 96, where v is the speed of the cyclist.

The cyclist overtakes the bus when d_bus = d_cyclist. So we can set the two equations equal to each other and solve for t:

1/2 * a * t^2 = v * t + 96

Substituting the given values:

1/2 * 2 * t^2 = 20 * t + 96

Solving this equation gives t = 8 seconds and t = -12 seconds. Since time cannot be negative, the cyclist will overtake the bus after 8 seconds. So, the correct answer is 8sec.