Answer:
Van will stop after a distance of 183.216 ft
Explanation:
We have given initial speed of the van = 36 mi/hr
As the van finally stops so final velocity v = 0 mi/hr
Distance after which van stop = 50 ft
As 1 ft = 0.000189 mi
So 50 ft [tex]=50\times 0.000189=0.00945mi[/tex]
From third equation of motion [tex]v^2=u^2+2as[/tex]
[tex]0^2=36^2+2\times a\times 0.00945[/tex]
[tex]a=-68571.42mi/hr^2[/tex]
In second case u = 69 mi/hr
And acceleration is same
So [tex]0^2=69^2-2\times 68571.42\times s[/tex]
[tex]s=0.03471mi[/tex]
As 1 mi = 5280 ft
So [tex]0.0347mi=0.0347\times 5280=183.216ft[/tex]
