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 #6/21 on the paper 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 #6/21 on the paper 15. An airplane travels 1000 ft at a constant acceleration while taking off. If it s

Explanation To solve the problem we will use the following two kinematics equations. vf2=vi2+2ad .................(1) vf=vi+at........................(2) where, vi is the initial velocity of the rocket. vf is the final velocity of the rocket. a is the acceleration. t is the time interval. Given Data Initial velocity of the airplane is vi=75.0 knots Final velocity of the rocket is vf=0 (as it finally comes to rest) Displacement of the airplane before it comes to rest is d=400 ft 0.869 knot=1.47 ft/sCalculation First we will convert the initial velocity in ft/s. It is given that, 0.869 knot=1.47 ft/s ⇒1 knot=1.470.869 ft/s ⇒1 knot=1.69 ft/s Now, vi=75.0 knots =75.0 knots×1.69 ft /s1 knot ⇒vi=127 ft/s Calculation for average acceleration substituting the values of vi, vf, a and d in equation 1 we get, 0=(127)2+2×a×400 ⇒16129+800a=0 ⇒800a=-16129 ⇒a=-16129800 ft/s2 ⇒a=-20.2 ft/s2 Answer Average acceleration of the airplane is -20.2 ft/s2 Calculation for the time Let after time t the airplane comes to rest. Now substituting the values of vf, vi and a in equation 2 we get, 0=127+(-20.2×t) ⇒127-20.2t =0 ⇒20.2t=127 ⇒t=12720.2 s ⇒t=6.29 s Answer So it takes 6.29 s by the plane to stop.