Phys 201 1st EditionFinal Exam Study Guide Lectures: 1 -24 Lecture 1 (August 27th)A commuting student leaves home and drives to school at an average speed of 40.0 km/h. After 20.0 min he realizes that he has forgotten his homework and returns home to get it at the same average speed. It takes 9.0 min to find the report, after which the trip to school 40.0 km away tothe east is resumed at the same speed as before. a) What is the average speed for the entire trip? b) What is the average velocity for the entire trip?Answer:a) 10.2m/sb) 6.12 m/sExplanation: The first thing we need to do is convert km/h into m/s for all variables. 40km/h(1000/3600m/s)= 11.11m/s20 minutes= 1200s9 minutes= 540s40km=40000ma) The speed is the total distance divided by time. To find the total distance, we add the first triptowards the school, the trip back home, and the last trip to the school.11.11m/s(1200s)= 13332m13332(2)=26664mTotal distance= 26664m+40000m=66664mTo get the speed we divided this distance by the total time traveled. To get the total time, we need to add the 20 minutes of the first two trips, the 9 minutes spend at home, and the hour to get back to school.66664m/(1200s+1200s+540s+3600s)=10.19m/sb) Velocity is the change in distance or displacement over the total time. This means that we only need to consider the distance from the house to the school divided by the total time. 40000m/(1200s+1200s+540s+3600s)=6.116m/s Lecture 2 (September 3rd)A cannonball is dropped from the top of a building. If the point of release is 42.6 m above the ground, what is the speed of the cannonball just before it strikes the ground?Answer: 28.9m/sExplanation: Because the ball is dropping straight down, it has a constant gravitational acceleration of 9.81m/s2. Also, the initial velocity of the ball at the top of the building is 0m/s. Therefore, to find the final velocity (speed) of the cannon ball, we use the following equations;X=V0t + .5at2V-V0=atWe use the first equation to solve for time, which we then plug into the second equation to solve for velocity.X=V0t + .5at2  42.6= 0t +.5(9.81)t2t= √(42.6/4.9) = 2.947sV-V0 =at  V-0=(9.81)(2.947)V=28.91m/sLecture 3(September 8th)A person desires to reach a point that is 5.20 km from her present location and in a direction that is 45° north of east. However, she must travel along streets that are oriented either north-south or east-west. What is the minimum distance she could travel to reach her destination? Answer: 7.35kmExplanation: For this problem we need to find the x and y components of the vector. We know that our vector is 5.2 at a 45◦ angle, so we can solve for the x and y components using our trig functions. Our equations would be sin(θ) = opp/hyp and cos(θ) = adj/hyp. Once we have the values of our x and y components, we add them together to find the minimum distance.Lecture 3(September 8th)A small airplane flies with a speed relative to the ground (ground speed) of 223.0 km/h in a direction 15.0 degrees to the east of north. If the plane is headed due north and the deviation from that direction is due to a cross wind blowing from west to east, what is the speed of the wind (in km/h)?Answer: 57.7km/hExplanation: If the wind is traveling west to east then we know that we are looking for the x component of the vector. The vector has a magnitude of 223 at 15◦. For this problem, the x component is adjacent to the 15◦ angle, so we will use the trigonometric equation, cos(θ) = adj/hyp. To find the x component, multiply 223*Cos(15).Lecture 4 (September 10th)A rock thrown horizontally from the top of a radio tower lands 24.0 m from the base of the tower. If the speed at which the object was projected was 10.50 m/s, how high is the tower?Answer: 25.6mExplanation: For this problem, we need to solve for Y. For this we will use the equation (y-yo)=vyot-.5gt2. To use this equation, we need to solve for time. Because there is no angle of elevation, the velocity of 10.5m/s is in the x direction. Therefor we can use the equation (x-xo)=Vxot, which will come to t=24/10.5. We then plug this value into our y equation. Because weassume that the rock stops when it hits the ground, Vyo =0. Therefor our y distance is 0.5*9.81*2.292.Lecture 5(September 15th)A small steel ball bearing with a mass of 27.0 g is on a short compressed spring. When aimed vertically and suddenly released, the spring sends the bearing to a height of 1.45 m. Calculate the horizontal distance the ball would travel if the same spring were aimed 28.0 deg from the horizontal.Answer: 2.4mExplanation: First thing we have to do is find the velocity of the ball using conservation of energy (PE=KE). The initial velocity is the y component of the velocity of the ball at 28deg. Once we know the y component, we can solve for the distance covered by the x component.PE=KE  mgh=.5mv2Vy=√(gh/.5)  V=√(9.81*1.45/.5)=5.334m/sD=v2sin(2θ)/a  d=5.3342Sin(56)/9.81D=2.404Lecture 6(September 17th)A 55.8-kg high-school student hangs from an overhead bar with both hands. What is the tensionin each arm if the bar is gripped with both arms raised vertically overhead? Answer: 274NExplanation: Tension is simply force. Force is mass times acceleration. In this case, acceleration is due to gravity, and therefore is 9.81m/s2. Therefore the tension of both arms is T=(55.8)(9.81) =547.398N. To find the tension in each arm, just divide this force in half.Lecture 6(September 17th)Two horses pull horizontally on ropes attached to a tree stump. Each horse pulls with a force of magnitude F. If the resultant force (vector R) has the magnitude R = 1.51 F, what is the angle (in degrees) between the two ropes?Answer: 81.9degExplanation: Forces are vectors, so we can use vector addition to solve for this problem. If we think of F and R in terms of a triangle, F is the hypotenuse and R is the adjacent angle (in conjunction with our angle). Because we have 2 triangles, R is divided in half. To find our angle measure, we will used the equation Cos(θ)=(adj/hyp)/2. This comes out to Cos(θ)=1.51/2=.755. We then take the arcos of this valueto get the angle for one of the triangles. To get the angle between both ropes, we double this value.Lecture 8 (September 29th)During 0.19 s, a wheel rotates through an angle of 3.42 rad as a point on the periphery of the wheel moves with a constant speed of 2.21 m/s. What is the radius of the wheel? Answer: .123mExplanation: The radius is equal to velocity divided by angular