Let's solve the homework question step by step. ### Part (a): Free Body Diagram and Kinetic Diagram at Point BAt point B, the car is at the top of the bump. The forces acting on the car are: 1. The gravitational force (weight) acting downward, \( F_g = mg \). 2. The normal force \( N \) exerted by the road on the car, acting upward. Since the car is about to leave the surface, the normal force \( N \) will be zero. The only force providing the centripetal acceleration is the gravitational force. ### Part (b): Instant Speed at Point BTo find the speed at point B such that the car is about to leave the surface, we use the centripetal force equation. At the top of the bump, the centripetal force is provided by the gravitational force: \[ F_g = \frac{mv^2}{r} \] Where: - \( F_g = mg \) is the gravitational force. - \( m \) is the mass of the car. - \( v \) is the speed of the car at point B. - \( r \) is the radius of curvature of the bump. Setting the gravitational force equal to the centripetal force: \[ mg = \frac{mv^2}{r} \] Solving for \( v \): \[ g = \frac{v^2}{r} \] \[ v^2 = gr \] \[ v = \sqrt{gr} \] Given: - \( g = 9.8 \, \text{m/s}^2 \) - \( r = 20 \, \text{m} \) \[ v = \sqrt{9.8 \times 20} \] \[ v = \sqrt{196} \] \[ v = 14 \, \text{m/s} \] So, the instant speed of the car at point B is \( 14 \, \text{m/s} \). ### Part (c): Constant Acceleration from A to BTo find the constant acceleration needed to change the speed from \( 30 \, \text{m/s} \) at point A to \( 14 \, \text{m/s} \) at point B over a distance of \( 100 \, \text{m} \), we use the kinematic equation: \[ v_f^2 = v_i^2 + 2a d \] Where: - \( v_f = 14 \, \text{m/s} \) is the final speed at point B. - \( v_i = 30 \, \text{m/s} \) is the initial speed at point A. - \( a \) is the constant acceleration. - \( d = 100 \, \text{m} \) is the distance. Rearranging to solve for \( a \): \[ a = \frac{v_f^2 - v_i^2}{2d} \] Substituting the given values: \[ a = \frac{14^2 - 30^2}{2 \times 100} \] \[ a = \frac{196 - 900}{200} \] \[ a = \frac{-704}{200} \] \[ a = -3.52 \, \text{m/s}^2 \] So, the car needs to decelerate at a constant rate of \( -3.52 \, \text{m/s}^2 \) to change its speed from \( 30 \, \text{m/s} \) to \( 14 \, \text{m/s} \) over a distance of \( 100 \, \text{m} \).

Let's solve the homework question step by step. ### Part (a): Free Body Diagram and Kinetic Diagram at Point BAt point B, the car is at the top of the bump. The forces acting on the car are: 1. The gravitational force (weight) acting downward, $F_g = mg$. 2. The normal force $N$ exerted by the road on the car, acting upward. Since the car is about to leave the surface, the normal force $N$ will be zero. The only force providing the centripetal acceleration is the gravitational force. ### Part (b): Instant Speed at Point BTo find the speed at point B such that the car is about to leave the surface, we use the centripetal force equation. At the top of the bump, the centripetal force is provided by the gravitational force: $F_g = \frac{mv^2}{r}$ Where: - $F_g = mg$ is the gravitational force. - $m$ is the mass of the car. - $v$ is the speed of the car at point B. - $r$ is the radius of curvature of the bump. Setting the gravitational force equal to the centripetal force: $mg = \frac{mv^2}{r}$ Solving for $v$: $g = \frac{v^2}{r}$ $v^2 = gr$ $v = \sqrt{gr}$ Given: - $g = 9.8 \, \text{m/s}^2$ - $r = 20 \, \text{m}$ $v = \sqrt{9.8 \times 20}$ $v = \sqrt{196}$ $v = 14 \, \text{m/s}$ So, the instant speed of the car at point B is $14 \, \text{m/s}$. ### Part (c): Constant Acceleration from A to BTo find the constant acceleration needed to change the speed from $30 \, \text{m/s}$ at point A to $14 \, \text{m/s}$ at point B over a distance of $100 \, \text{m}$, we use the kinematic equation: $v_f^2 = v_i^2 + 2a d$ Where: - $v_f = 14 \, \text{m/s}$ is the final speed at point B. - $v_i = 30 \, \text{m/s}$ is the initial speed at point A. - $a$ is the constant acceleration. - $d = 100 \, \text{m}$ is the distance. Rearranging to solve for $a$: $a = \frac{v_f^2 - v_i^2}{2d}$ Substituting the given values: $a = \frac{14^2 - 30^2}{2 \times 100}$ $a = \frac{196 - 900}{200}$ $a = \frac{-704}{200}$ $a = -3.52 \, \text{m/s}^2$ So, the car needs to decelerate at a constant rate of $-3.52 \, \text{m/s}^2$ to change its speed from $30 \, \text{m/s}$ to $14 \, \text{m/s}$ over a distance of $100 \, \text{m}$.