Physics Solution Manual for 1100 and 2101

Solution the table below lists the scalar components

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: , we have y = –(v0 sin 30.0°) t + 1 a y t 2 2 which can be rearranged to give the following equation that is quadratic in t: 1 a t2 2y − (v0 sin 30.0°) t − y = 0 Using y = –2400 m and a y = −9.80 m/s 2 and suppressing the units, we obtain the quadratic equation 4.9t 2 + 120t − 2400 = 0 Using the quadratic formula, we obtain t = 13 s. Therefore, we find that x = ( v 0 cos 30 .0 ° ) t = ( 240 m / s ) (cos 30.0 ° ) ( 13 s) = 2700 m Equation (1) then gives 2400 m = 42° 2700 m θ = tan −1 76. REASONING The relative velocities in this problem are: vPW = velocity of the Passenger relative to the Water vPB = velocity of the Passenger relative to the Boat (2.50 m/s, due east) vBW = velocity of the Boat relative to the Water (5.50 m/s, at 38.0° north of east) The velocities are shown in the drawing are related by the subscripting method discussed in Section 3.4: vPW = vPB + vBW 154 KINEMATICS IN TWO DIMENSIONS +y (North) vPW vPW +x (East) 5.50 m/s vBW θ θ vPB 2.50 m/s 38.0° We will determine the magnitude and direction of vPW from the equation above by using the method of scalar components. SOLUTION The table below lists the scalar components of the three vectors. Vector x Component y Component vPB +2.50 m/s 0 m/s vBW vPW = vPB + vBW +(5.50 m/s) cos 38.0° = +4.33 m/s +(5.50 m/s) sin 38.0° = +3.39 m/s +2.50 m/s + 4.33 m/s = +6.83 m/s 0 m/s + 3.39 m/s = +3.39 m/s The magnitude of vPW can be found by applying the Pythagorean theorem to the x and y components: vPW = ( 6.83 m/s ) + ( 3.39 m/s ) 2 2 = 7.63 m/s The angle θ (see the drawings) that vPW makes with due east is +3.39 m/s = 26.4° north of east +6.83 m/s ______________________________________________________________________________ θ = tan −1 77. REASONING The drawings show the initial and final velocities of the ski jumper and their 2 2 scalar components. The initial speed of the ski jumper is given by v0 = v0 x + v0 y , and the v angle that the initial velocity makes with the horizontal is θ = tan −1 0 y . The scalar v 0x components v0x and v0y can be determined by using the equations of kinematics and the data in the following tables. (The +x direction is in the direction of the horizontal displacement of the skier, and the +y direction is “up.”) Chapter 3 Problems v0 155 vx v0y 43.0° θ v v0x Initial velocity vy Final velocity x-Direction Data x ax vx v0x +51.0 m 0 m/s2 +(23.0 m/s) cos 43.0° = +16.8 m/s ? t y-Direction Data y ay −9.80 m/s vy 2 v0y −(23.0 m/s) sin 43.0° = −15.7 m/s t ? Since there is no acceleration in the x direction (ax = 0 m/s2), v0x is the same as vx, so we have that v0x = vx = +16.8 m/s. The time that the skier is in the air can be found from the x-direction data, since three of the variables are known. With the value for the time and the y-direction data, the y component of the initial velocity can be determined. 2 SOLUTION Since ax = 0 m/s , the time can be determined from Equation 3.5a as x +51.0 m = = 3.04 s . The value for v0y can now be found by using Equation 3.3b t= v0 x +16.8 m/s with this value of the time and the y-direction data: v0 y = v y − a y t = −15.7 m/s − ( −9.80 m/s 2 ) ( 3.04 s ) = +14.1 m/s The speed of the skier when he leaves the end of the ramp is 2 2 v0 = v0 x + v0 y = ( +16.8 m/s ) + ( +14.1 m/s ) 2 2 = 21.9 m/s The angle that the initial velocity makes with respect to the horizontal is v0 y −1 +14.1 m/s = tan = 40.0° v0 x +16.8 m/s ______________________________________________________________________________ θ = tan −1 156 KINEMATICS IN TWO DIMENSIONS 78. REASONING The relative velocities in this problem are: vPS = velocity of the Passenger relative to the Shore vP2 = velocity of the Passenger relative to Boat 2 (1.20 m/s, due east) v2S = velocity of Boat 2 relative to the Shore v21 = velocity of Boat 2 relative to Boat 1 (1.60 m/s, at 30.0° north of east) v1S = velocity of Boat 1 relative to the Shore (3.00 m/s, due north) The velocity vPS of the passenger relative to the shore is related to vP2 and v2S by (see the method of subscripting discussed in Section 3.4): vPS = vP2 + v2S But v2S, the velocity of Boat 2 relative to the shore, is related to v21 and v1S by +y (North) v2S = v21 + v1S v1S vPS 3.00 m/s Substituting this expression for v2S into the first equation yields vPS = vP2 + v21 + v1S 1.60 m/s This vector sum is shown in the diagram. We will determine the magnitude of vPS from the equation above by using the method of scalar components. 1.20 m/s 30.0° v21 +x (East) vP2 SOLUTION The table below lists the scalar components of the four vectors in the drawing. Vector x Component y Component vP2 +1.20 m/s 0 m/s v21 v1S vPS = vP2 + v21 + v1S +(1.60 m/s) cos 30.0° = +1.39 m/s +(1.60 m/s) sin 30.0° = +0.80 m/s 0 m/s +3.00 m/s +1.20 m/s + 1.39 m/s = +2.59 m/s +0.80 m/s +3.00 m/s = +3.80 m/s The magnitude of vPS can be found by applying the Pythagorean theorem to its x and y components: vPS = ( 2.59 m/s ) + ( 3.80 m/s ) = 4.60 m/s _______________________________...
View Full Document

This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

Ask a homework question - tutors are online