v = d = df – di . a = v = vf – vi .t t f – t I t t f – tivf = vi + a t v = ½ ( vf + vi )d = ½ ( vf + vi ) t d = vi t + ½ a t2vf2 = vi2 + 2 a d #6 / 21on the page ### Dynamics and Kinematics Problems#### 1. Airplane Takeoff Velocity**Problem 15:**- An airplane travels 1000 ft at a constant acceleration while taking off. If it starts from rest and takes off in 25.0 s, what is its takeoff velocity?#### 2. Car Velocity and Position**Problem 17:**- A car starts from rest and acquires a velocity of 20.0 mph in 10.0 s. Determine its location and velocity at 10.0, 15.0, 20.0, and 25.0 s.#### 3. Electron Acceleration**Problem 19:**- An electron in a vacuum tube acquires a velocity of \(5.3 \times 10^8\) cm/s over a distance of 0.25 cm. Find the acceleration of the electron.#### 4. Airplane Deceleration**Problem 21:**- An airplane has a touchdown velocity of 75.0 knots and comes to rest in 400 ft. Calculate the airplane's average deceleration and the time taken to stop. - Note: \(0.869 \text{ knot} = 1.47 \text{ ft/s}\)#### 5. Car Stopping Distance**Problem 23:**- The speedometer of a car reads 60.0 mph when the brakes are applied. The car comes to rest in 4.55 s. Calculate the distance traveled before coming to rest.#### 6. Reaction Time and Stopping Distance**Problem 25:**- A driver traveling at 25.0 mph notices a red light at an intersection. With a reaction time of 0.600 s and a car deceleration of \(18.0 \text{ ft/s}^2\), find the stopping distance. What would the stopping distance be if the car were moving at 50.0 mph?

Name: v = d = df – di . a = v = vf – vi .t t f – t I t t f – tivf = vi +
Uploaded: 2024-03-05T11:56:34.000Z
Channel: Bartleby
Description: v = d = df – di . a = v = vf – vi .t t f – t I t t f – tivf = vi + a t v = ½ ( vf + vi )d = ½ ( vf + vi ) t d = vi t + ½ a t2vf2 = vi2 + 2 a d #6 / 21on the page ### Dynamics and Kinematics Problems#### 1. Airplane Takeoff Velocity**Problem 15:**