Final answer:
The speed of the roller coaster at the top of a loop with a 15.0 m radius of curvature and 1.50 g downward acceleration can be found using the centripetal acceleration formula by solving for the velocity that ensures passengers stay firmly in their seats due to sufficient centripetal force overcoming gravity.
Step-by-step explanation:
The question pertains to the physics of roller coasters and specifically addresses a scenario involving centripetal force and acceleration in the context of a roller coaster loop. When a roller coaster reaches the top of a loop, the centripetal acceleration needs to be strong enough to overcome gravity and keep the passengers pressed in their seats. In the given scenario, the coaster at the top of the loop has a radius of curvature of 15.0 m and a downward acceleration of 1.50 g (where g is the acceleration due to gravity, approximately 9.8 m/s2). To maintain this acceleration, we would use the formula for centripetal acceleration (a = v2 / r) where a is acceleration, v is velocity, and r is the radius of curvature.
The speed at the top of the loop can be found by rearranging this formula to v = √(a × r). We multiply the standard acceleration due to gravity, 9.8 m/s2, by 1.50 to get the downward acceleration, and then multiply by the radius (15.0 m), before taking the square root to solve for the velocity. This will give us the minimum speed needed at the top of the loop to keep the riders safely in their seats. Engineers use such principles to design safe and exciting roller coasters.