Answer:
The distance traveled by car before stopping is [tex]90\, ft[/tex]
Explanation:
Given
Initial speed of car [tex]u=30\, \frac{ft}{sec}[/tex]
Final speed of the car [tex]v=0\, \frac{ft}{sec}[/tex]
Acceleration of the car [tex]a=-5\, \frac{ft}{sec^{2}}[/tex]
Let distance traveled before stopping be s
Using equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
=>[tex]0^{2}=30^{2}+2\times (-5)\times s[/tex]
=>[tex]s=90\, ft[/tex]
Thus the distance traveled by car before stopping is [tex]90\, ft[/tex]
