A speedboat has a velocity of 60 mph at A and accelerates at

A speedboat has a velocity of 60 mph at A and accelerates at a constant rate of 1 m/s^2 between points A and B following the dashed line as shown in Figure 2. Assuming that the speedboat travels 145 m to reach point B and that the radius of curvature at that point is 130 m. determine: the speed of the speedboat at point B and the time it takes to reach point B; the force vector exerted on the driver at point B assuming the mass of the driver is 70 kg (express the vector using the i and j unit vector coordinates). An aircraft lands in a carrier as shown in Figure 3. Determine the distance it will travel on the surface of the carrier before completely stopping. Assume that: the mass of the aircraft is 10.6 metric tons; at A, the height of the aircraft is 25 m and the corresponding speed is 125 km/h; the deck arresting cable (that will aid the aircraft to stop) has a spring constant of 40 kN/m and will activate 20 m after the aircraft touches the carrier surface; the friction force is 50 N/m.

Solution

Given data

VA = 60mph

a = 1m/s²

S = 145m

R = 130m

VB = ?

Fvector = ?

VB will be found by using third equation of motion

          2as = Vf² - Vi²

           2as = VB² - VA²

          V_B² = 2as + VA²

                 = 2(1)(145) + 26.8                  ( 60mph = 26.8 m/s)

                  = 316.8

              VB = 17.79 m/s

               VB = 39.79mph

This would be the value of velocity at point B

       At point B the force would be the centripetal force that is the total net foce would be acting to the center of the curvature and its value will be

               F= m V_B² / R

                VB(vector) = V_B cos (360-20) i + V_B sin (260 -20) j

                V_B (vecotor) = 39.9 mph ( cos340⁰ i + sin 340 j)           Angles are taken relative to the positive x axis.

Finding the unit vector of velocity

                         V(unit vector ) = cos340 i + sin 340 j

                                               = 0.93i - 0.34j

           F vector =( mV_B² / R) (0.39i - 0.34j)

                         = (70 ( 17.79)² / 130 ) ( 0.39 i - 0.34j)

                        = ( 66.41i - 57.93 j ) N

            This would be the force vector at point B