Question

1. free - fall ride suppose a free - fall ride at an amusement park starts at rest and is in free fall. what is the velocity of the ride after 2.3 s? how far do people on the ride fall during the 2.3 - s time period? science notebook - accelerated motion 53

Explanation:

Step1: Identify the relevant kinematic - equation

For free - fall motion, the initial velocity \(u = 0\ m/s\) (starts at rest), the acceleration \(a = g=9.8\ m/s^{2}\), and we want to find the velocity \(v\) and the displacement \(s\) at time \(t = 2.3\ s\). The velocity - time equation is \(v=u + at\) and the displacement - time equation is \(s=ut+\frac{1}{2}at^{2}\).

Step2: Calculate the velocity

Since \(u = 0\ m/s\), \(a = g = 9.8\ m/s^{2}\), and \(t = 2.3\ s\), using the formula \(v=u + at\), we substitute the values: \(v=0+9.8\times2.3\).

$$v = 9.8\times2.3=22.54\ m/s$$

Step3: Calculate the displacement

Since \(u = 0\ m/s\), \(a = g = 9.8\ m/s^{2}\), and \(t = 2.3\ s\), using the formula \(s=ut+\frac{1}{2}at^{2}\), we substitute the values: \(s = 0\times t+\frac{1}{2}\times9.8\times(2.3)^{2}\).

$$s=\frac{1}{2}\times9.8\times(2.3)^{2}=4.9\times5.29 = 25.921\ m$$

Answer:

The velocity of the ride after \(2.3\ s\) is \(22.54\ m/s\) and the distance people on the ride fall during the \(2.3 - s\) time period is \(25.921\ m\)