Better Safe Than Sorry

Firstly, $90 \mathrm{ km}/\mathrm{h} = 25 \mathrm{ m}/\mathrm{s}$ . We will use the equation $v^2=u^2+2ax$ , with $v=0$ , $u=25$ and $a=-1.25$ to determine how far it travels before it stops moving. From this, we get that $0=625-2.5x$ , which implies $x=250 \mathrm{ m}$ . Thus, the car has travelled 250 metres before it stops. But the sinkhole is only 200 metres. Thus, the car will fall into the sinkhole.

$u=$ $90 km/h$ = $90$ $*$ $\frac{5}{18}$ $m/s$ $=$ $25 m/s$

$v=$ $0$ $m/s$

$a=$ $- 1.25$ $m/s^{2}$

Using the equation of motion : $v^2=u^2+2as$ → $v^2 - u^2 = 2as$ → $s=$ $\frac{v^2 - u^2}{2a}$

Putting the values, $s =$ $\frac{625-0}{2.5}$ → $s \ = \ 250 m$ (50 m more than where the car should have stopped to get saved.)

So the car will fall into the sinkhole.

90 km/h = 25m/s.
Let $x$ = time in seconds, and let the function $f(x)$ = the speed of the vehicle at time x.
If the vehicle is traveling at 25m/s, applies the breaks at $x=0$ , and decelerates at a rate of $-1.25m/s^2$ ,
Then $f(x)=25-1.25x$ , and $f(20)=0$ (speed reaches zero at 20 seconds).
if $f(x)$ is our speed at time x, then $F(x)$ is our total distance traveled at time x.
so $F(x) = \int {f(x)dx} = 25x - .625x^2$ .
We know the speed reaches zero at 20 seconds, so $F(20) = 250$ (vehicle travels 250 meters before stopping).
Therefore it falls in the hole.
Additionally, we can solve for x in $F(x)=200$ and see that the vehicle reaches the hole at x = 11.056 seconds, and
$f(11.056) = 11.18$ shows us that the vehicle is traveling at 11.18 m/s upon reaching the hole (and thus will fall in).

Use the v2 - u2 = 2as

S is found to be 250 m... Which is more than 200 m thats why the car will fall into the sink hole