Man vs Bus

Suppose the bus and the man are moving in the positive $x$ direction. At time $t = 0$ , let the man be at $x = 0$ , and the bus be at $x = b$ . Then the equations of motion of the bus and the man are $x_b(t) = 10t + b$ , and $x_m(t) = \frac{1}{2} \times 2 \times t^2$ respectively.

For the man to be able to catch the bus, the equation $x_m - x_b$ should have a solution for some $t \ge 0$ .

$\begin{aligned} x_m - x_b &= t^2 - 10t + b\\ & = (t^2 - 2 \times 5 t + 25) - 25 + b\\ & = (t - 5)^2 + (b - 25)\\ \end{aligned}$

We see that this is a parabola, with minimum value $b - 25$ . If $b \le 25$ , then $x_m - x_b$ cuts the $x$ axis at least one, and has a solution for $t \ge 0$ . Hence the initial distance between the man and the bus should be no greater than 25 m.

Here,

The previous velocity of the bus, $V_{0} = \SI[per-mode=symbol]{0}{\meter\per\second}.$ . The acceleration of the bus, $a =\SI[per-mode=symbol]{2}{\meter\per\second\squared}.$ The uniform velocity of that man, $V =\SI[per-mode=symbol]{10}{\meter\per\second}.$ .

Suppose,

After $t$ seconds the man caught that bus,at that moment the bus has crossed $s$ distance. Then for Bus now we can get,

$S=V_{0}t+\dfrac{1}{2}at^{2}$

$or,S=0+\dfrac{1}{2}\times 2\times t^{2}$

$s=t^{2}$ ..........................................(a)

Again Suppose that,

frrom $x$ m behind the man run after the bus.So, to catch the bus he needs to cross $(s+x)$ distance at the time $t$ .

So, for that man now we can get,

$S+x=Vt$

$or,S+x=10t$ .........................(b)

By doing $(b-a)$ now we can get,

$x=10t-t^{2}$

$or,t^{2}-10t+x=0$ ...........(c)

Here, if the value of $t$ is real then the man can catch the bus,

So, by applying this method on equation... $(c)$ which is used for real value

$b^{2}-4ac\geq 0$

$(10)^{2}-4x\geq 0$

$or,-4x\geq-100$

$or,4x\leq100$

$or,x\leq25$

So, The man can reach the bus if the initial distance between them is $25$ m or less than $25$ m.

Assume t seconds after the start the bus and the man are at the same place and bus' speed just touched 10m/s. Which means after t seconds he cannot catch the bus because it is accelerating. Since we have the value for acceleration, using the equation v=at we can calculate the t.

10 m/s = t × 2 m/s^2 So t = 5 seconds

From equation S = 1/2 × a × t^2 we can deduct the bus travelled 25 mtrs in 5 seconds. In the same 5 seconds the man would have travelled 10 × 5 = 50 mtrs. The difference of 25 mtrs hence is the distance between the bus and the man at the start and hence the limit. As long as the distance is <= 25 mtrs, the man can catch the bus.