Question

A bus starts moving with acceleration 2 m/s^2. a cyclist 96m behind the bus starts simultaneously towards the bus at 20 m/s . after what time will he be able to overtake the bus.

Answer 1

Solutions 

We know that a bus starts moving with acceleration 2 m/s² and a cyclist 96m behind the bus starts biking simultaneously towards the bus at 20 m/s. 

Lets suppose the position of the cyclist to be "0" (t= 0) 

The position of the bus is 96 + $\frac{1}{2}$ a t² = 96 + t²  

The position of the cyclist is Vt = 20 t 

Therefore where they will meet ⇒ 20 t = 96 + t² 

To solve for variable t we have to use the standard formula for quadratics. 

t² - 20 t + 96 = 0 

⇒ (20 +- 4) /2 

The cyclist passes the bus at (20 - 4) /2 = 8 s 

The bus overtakes the cyclist at (20+4)/2 = 12 s