My Notes Aak Your Teacher 4. 0/2 points | Previous Answers SerCP11 3.P015. A car

Question

My Notes Aak Your Teacher 4. 0/2 points | Previous Answers SerCP11 3.P015. A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 21.0° below the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline with a constant acceleration of 3.37 m/s? for a distance of 35.0 m to the edge of the cliff, which is 40.0 m above the ocean. (a) Find the car's position relative to the base of the cliff when the car lands in the ocean. (b) Find the length of time the car is in the air. Your response differs from the correct answer by more than 10%. Double check your calculations. s Need Help? Read It

Explanation / Answer

The speed of the car when it reaches the edge of the cliff is

v2 = u2 + 2 a s = 0 + 2 x 3.37 x 35 = 235.9

v = 15.36 m/s

The horizontal velocity vox = 15.36 x cos 21° = 14.34 m/s

Vertical velocity vyo = -15.36 x sin 21 = -5.5 m/s

(b) The length of time the car is in the air

y = yo + vyo x t + (1/2) ay x t2

y = yo + vyo * t - (1/2) g t2

0 = 40 + ( - 5.5) t - (1/2) (9.8 ) t2

4.9t2 + 5.5 t - 30 = 0

Solve this quadratic equation you get t = 1.98 sec

(a) position of the car when it lands in the ocean, relative to the base of the cliff.

x = vox x t = 14.34 x 1.98 = 28.34 m

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