Question

the minimum distance required to stop a car moving at 35 mi/h is 42.6 ft what is the minimum stopping distance for the same car moving at 70 mi/h

Answer 1

Given:

$v_1=35\: \rm mi/h$

$d_1=\rm 42.6 \: ft$

$v_2=70\: {\rm mi/h}=2v_1$

The energy-work theorem says

$$\frac{mv^2}{2}=Fd$$

Hence, the stopping distance

$$d\sim v^2$$

So

$$d_1\sim v_1^2\\ d_2\sim v_2^2=(2v_1)^2=4d_1\\=4*42.6\rm \:ft=170.4\: ft$$