A car is parked on a steep incline, making an angle of 37.0 degrees below the ho

Question

A car is parked on a steep incline, making an angle of 37.0 degrees below the horizontal and overlooking the ocean, when its breaks fail and it begins to roll. Starting from rest at t = 0, the car rolls down the incline with a constant acceleration of 4.00 m/s^2, traveling 50.0 m to the edge of a vertical cliff. The cliff is 30.0m above the ocean. Find (a) the speed of the car when it reaches the edge of the cliff, (b) the time interval clapsed when it arrives there, (c) the velocity of the car when it lands in the ocean, (d) the total time interval the car is in motion, and (e) the position of the car when it lands in the ocean, relative to the base of the cliff.

Explanation / Answer

1) 55 = 0t + 1/2at^2 and you know a solve for t t = 6.63 s v = 0 + at = (2.5)(6.63) = 16.6 m/s vinitial = 16.6 m/s vxinitial = 16.6cos-37 = 13.3 m/s vyiniital = 16.6sin-37 = -10.0m/s vyfinal^2 = vyinitial^2 - 2gy vyfinal = 26.2 m/s vxfinal = 13.3 m/s vfinal = sqrt(vx^2+vy^2) = 29.4 m/s direction = 63.1 degrees below horizontal time1 = 6.63 s time2: 26.2 =10+(9.8)t time2 = 1.65 s total time = 8.28 seconds x = vxt = 13.3(1.65) = 21.9 meters from base 2) Use the equations same as above and make sure you break things up into components. 3) Okay first convert the km/h to m/s. There are two accelerations Centripetal and linear Linear = vf - vi / t Centrip. = v^2 / r Treat these as two separate vectors. a = sq rt.(alinear^2 + acentrip.^2) Direction = taninverse(acentrip / alinear)

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