Given:
The distance covered by the car after the driver applies the brake, d=119.7 m
The acceleration of the car, a=-10.94 m/s²
The final velocity of the car, v=0 m/s
To find:
The time it takes for the car to come to stop.
Explanation:
From the equation of motion,
[tex]d=vt-\frac{1}{2}at^2[/tex]On substituting the known values,
[tex]\begin{gathered} 119.7=0-\frac{1}{2}\times-10.94\times t^2 \\ t=\sqrt{\frac{119.7\times2}{10.94}} \\ =4.7\text{ s} \end{gathered}[/tex]Final answer:
Thus the car comes to rest in 4.7 s
