Question

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

Answer 1

Let the distance covered by Cyclist before meeting the bus be x metres.
then, total distance covered by bus when it will meet the cyclist will be (x-96) metres.

$s = ut + \frac{1}{2} a {t}^{2} \\ x - 96 = 0 + \frac{1}{2 } \times 2 \times {t}^{2} \\ {t}^{2} = x - 96 \\ t = \sqrt{x - 96} ....................(1)$

And now,

$time = \frac{distance}{speed} \\ t = \frac{x}{20} ................(2)$

Equalising (1) & (2)

$\sqrt{x - 96} = \frac{x}{20} \\ 20 \sqrt{x - 96} = x \\ x = 20 \times {(x - 96)}^{ \frac{1}{2} } \\x = {400}^{ \frac{1}{2} } \times {(x - 96)}^{ \frac{1}{2} } \\ x ={(400x - 38400)}^{ \frac{1}{2} } \\ {x}^{2} = 400x - 38400 \\ {x}^{2} - 400x + 38400 = 0 \\ {x}^{2} - 160x - 240x + 38400 = 0 \\ x(x - 160) - 240(x - 160) = 0 \\ (x - 160)(x - 240) = 0 \\ x = 160 \: or \: x = 240$

putting the value of x in eqn (2)

$t = \frac{x}{20} \\ t = \frac{160 \: and \: 240}{20} \\ t = 8 \: and \: 12 \: seconds.$

Cyclist will be able to overtake bus in 8 second after the start.
&
The will able to overtake Cyclist in 12 second after the start.

Original Answer will be 8 seconds because they will again cross each other at 12 seconds.

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