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 #7/23 on the page 15. An airplane travels 1000 ft at a constant acceleration while taking off. If it starts from rest, andtakes off in 25.0 s, what is its takeoff velocity?2E17. A car starts from rest and acquires a velocity of 20,0 mph in 10.0 s. Where is the car located andwhat is its velocity at 10.0, 15.0, 20.0, and 25.0 s?19. An electron in a vacuum tube acquires a velocity of 5.3 X 108 cmls in a distance of 0.25 cm. Find theacceleration of the electron.(6)21. An airplane has a touchdown velocity of 75.0 knots and comes to rest in 400 ft. What is theairplane's average deceleration? How long does it take the plane to stop? [0.869 knot = 1.47ft/s]23. The speedometer of a car reads 60.0 mph when the brakes are applied. The car comes to rest in4.55 s. How far does the car travel before coming to rest?25. A driver traveling at 25.0 mph sees the light turn red at the intersection. If her reaction time is0.600 s, and the car can decelerate at 18.0 ft/s, find the stopping distance of the car. Whatwould the stopping distance be if the car were moving at 50.0 mph?11.0.300 m13. 1.86 ft/s2; 475 f15. 54.5 mph (or 8017. t 10 s: d= 147t= 15 s: d= 33t3D= 20 s: d= 58t= 25 s: d= 9119.5.62 X 107 cm/21. -20.2 ft/s; 6.29 s23. 200 ft25. 59.4 ft; 193 ft

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 #7/23 on the page 15. An airplane travels 1000 ft at a constant acceleration while taking off. If it s