Phys 201 1nd EditionExam # 1 Study Guide Lectures: 1 - 6Lecture 1 (August 27th)A helicopter leaves its base and travels 22.0 km north. After a brief stop it flies 34.9 km south, pauses briefly and then flies 17.1 km north. Finally it flies 10.0 km south and lands. At the end of the trip, what is the displacement of the helicopter from its base?Answer: -5.8kmExplanation: For this problem we are solving for displacement in one-dimensional motion. This means that all we have to do is subtract the distance that the helicopter flies south from the distance that it fliesnorth. (22+17.1)-(34.9+10)= -5.8. Because the displacement is negative, we know the helicopter is 5.8kmsouth of his starting point.Lecture 2(September 3rd)A soccer ball is released from the top of a smooth incline. After 3.86 s the ball travels 9.0 m. One second later it has reached the bottom of the incline. A.) Assume the ball's acceleration is constant and determine its value ( m/s2). B) How long is the incline?Answer: A) 1.208 m/s2B) 14.29mExplanation: Our equations for motion with a constant acceleration are v = v0 + at and (x −x0) = v0t +.5at2. To solve this problem, we will divide our incline into two parts; the whole incline and the known incline. For this problem we can use the equation (x −x0) = v0t +.5at2. Because our acceleration is constant, we can find the acceleration by solving for our smaller triangle. In this case, (x-xo)=9, t=3.86, and Vo=0.This makes our equation a=(x-xo)/.5t2  a=9/(14.99*0.5)= 1.21m/s2. Now that we know the acceleration, we can solve for our big triangle using the same equation to solve for (x-xo). The ball reaches the end of the incline 1 second after it reaches the bottom of the smaller triangle, making t=4.86. This means (x-xo)=.5(1.21)(4.862 )=14.29m.Lecture 2(September 3rd)A motorcycle rider moving with an initial velocity of 6.4 m/s uniformly accelerates to a speed of 18.1 m/sin a distance of 32.0 m. A) What is the acceleration? B) How long does it take him to travel this distance?Answer: A)4.8m/s2B) 2.61sExplanation: To find acceleration without a value for time, we can use the equation v2= v20 + 2a(x − x0). We solve this as a=(18.12-6.42)/(2*32). To find the amount of time required to travel this distance, all we have to do is plug our value for acceleration into the equation v = v0 + at, meaning we would solve it as t=(18.1-6.4)/4.8Lecture 2(September 3rd)A car accelerates uniformly and reaches a speed of 4.00 m/s in 8.00 seconds. Calculate the distance traveled by the car from a time of 1.90 to 4.00 seconds.Answer: 3.1mExplanation: To find this distance, we can use the equation x − x0 =.5at2. To find the acceleration we just divide 4m/s by 8s. To get time we subtract 1.9 from 4. Then we just plug in our values to find the distance.Lecture 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 tothe 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 we assume 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) How far will a stone travel over level ground if it is thrown upward at an angle of 32.0 o with respect tothe horizontal and with a speed of 14.0 m/s? B) What is the maximum range that could be achieved withthe same initial speed?Answer: A)18mB) 20mExplanation: For the first part of this problem we are trying to find the x distance. For this we can use therange equation, R=(Vo2/g)Sin(2θ). The range that we get is the x distance. To get the maximum range, we will use the same equation, except we will plug in 45◦ for θ. This is because the maximum value of Sin comes from Sin(90◦), which is equal to 1.Lecture 5(September 15th)A Hollywood daredevil plans to jump the canyon shown in the figure on a motorcycle. There is a 15. m drop and the horizontal distance originally planned was 60. m but it turns out the canyon is really 66.0 macross.A) If he desires a 3.2-second flight time, what is the correct angle for his launch ramp?B) What is his correct launch speed?C) What is the correct angle for his landing ramp?D) What is his predicted landing velocity?Answer: A) 28.1◦B) 23.4 m/sC) 44.7◦D) 29m/sExplanation: The information about the original distance that the canyon was supposed to be is irrelevant. To solve for the launch angle, we need to solve for the Vy and Vx components. To solve for Vy we will use the equation y − y0= Vy0t − .5gt2. The Y distance is the height of the canyon. The x distance is the length (66m) of the canyon. We can plug this value into the equation x-xo=Vxot. Once we have our values for Vx and Vy, we can plug them into the