Answer:
It will take [tex]t=16.66s[/tex] for the lead car to stop.
Explanation:
Stage 1: The car moves at constant speed [tex]v_{i}=25\frac{m}{s}[/tex].
Stage 2: the car accelerates with [tex]a=-1.5\frac{m}{s^{2} }[/tex] (the negative sign means it's decelerating with respect to the direction is moving).
¿When will be the velocity be [tex]v=0[/tex]?
From Kinematics: [tex]v(t)=v_{0}+at[/tex]
So, if final velocity is [tex]v(t^{*} )=0[/tex]
then,
[tex]0=v_{0}+at^{*}[/tex] ⇒ [tex]t^{*}=\frac{-v_{0}}{a}[/tex] ⇒ [tex]t^{*}=16.66s[/tex].