Answer:
12.12 m/s
Explanation:
The net force of the car at the bottom of the dip is equal to
[tex]\begin{gathered} F__{net}=F_{seat}-mg=ma_c \\ F_n-mg=m\frac{v^2}{r} \end{gathered}[/tex]Where Fn is the normal force that acts upwards, m is the mass of the car, g is the gravity, v is the car's speed at the bottom of the dip and r is the radius.
We know that the weight of the passenger increase by 50%, so
Fn = 1.5mg
Replacing this on the first equation and solving for the car speed v, we get
[tex]\begin{gathered} 1.5mg-mg=\frac{mv^2}{r} \\ 0.5mg=\frac{mv^2}{r} \\ 0.5g=\frac{v^2}{r} \\ 0.5gr=v^2 \\ v=\sqrt{0.5gr} \end{gathered}[/tex]Finally, replacing g = 9.8 m/s² and r = 30 m, we get:
[tex]v=\sqrt{0.5(9.8)(30)}=12.12\text{ m/s}[/tex]So, the car’s speed at the bottom of the dip is 12.12 m/s
