Respuesta :
Given,
The initial velocity of the car, u=13.6 m/s
The final velocity of the car, v=0 m/s
The distance covered by the car, d=0.321 m
From the equation of the motion,
[tex]v^2-u^2=2ad[/tex]On rearranging the equation,
[tex]a=\frac{v^2-u^2}{2d}[/tex]This is the acceleration of the car which brings the car to rest after the collision.
On substituting the known values,
[tex]\begin{gathered} a=\frac{0-13.6^2}{2\times0.321} \\ =-288m/s^2 \end{gathered}[/tex]From another equation of the motion,
[tex]v=u+at[/tex]On rearranging the above equation,
[tex]t=\frac{v-u}{a}[/tex]On substituting the known values,
[tex]\begin{gathered} t=\frac{0-13.6}{-288} \\ =0.0472\text{ s} \end{gathered}[/tex]Thus the car comes to stop in 0.0472 seconds.
