Question

this is very hard plzz solve it i will mark it as brainlist

Answer 1

Given

$speed = 72 \frac{km}{hr} \\ \\ speed = 72 \times \frac{1000 \: m}{3600 \: s} \\ \\ speed = 20 \: \frac{m}{s}$

Given that it takes 10m distance to stop.

We have the formula

${v}^{2} - {u}^{2} = 2as \\ \\ final \: speed \: is \: 0 \\ initial \: speed \: is \: 20 \frac{m}{s} \\ and \: s \: is \: 10m \\ \\ so \: acceleration \: becomes \: a = \frac{{20}^{2}}{2 \times 10} \\ \\ so \: acceleration \: is \: 20 \frac{m}{s}$
a = 20m/s^2

Now
$speed \: = 144 \frac{km}{hr} \\ \\ = 144 \times \frac{1000m}{3600s} \\ \\ = 40 \frac{m}{s} \\ \\ now \: by \: the \: formula \: we \: get \: s = \frac{ {v}^{2} }{2a} \\ \\ s = \frac{ 40 \times 40}{2 \times 20} \\ \\ so \: s = 40m$

So the distance it travels before stopping is 40m