MCAT Physics - Problem Drill 06: Translational Motion Question No. 1 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 1. An object falls from rest from a high building. It falls for some time, t, and coers a distance, d. If a similar object falls for time, t, how far will it fall? Question #01 (A) d/ (B) d (C) d (D) 4d A. Incorrect! A greater time falling would lead to at least somewhat more distance coered than with time t. B. Incorrect! A greater time falling would lead to at least somewhat more distance coered than with time t. C. Incorrect! This answer may seem reasonable, but the distance coered depends on t squared, not just t. D. Correct! The distance coered depends on t squared. Since, d=it+at /, when the time is doubled, the distance coered is quadrupled. Therefore 4d is the correct answer. Consider your kinematics formulas. Specifically d = it+at /. Since the distance depends on time squared. Doubling the time will quadruple the distance since is 4. (D) is the correct answer.
Question No. of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as. A mile is approximately 1609m. Calculate the aerage speed in m/s of a quick runner that coers one mile in exactly 4 minutes. Question #0 (A) 6.7 m/s (B) 6.8 m/s (C) 40 m/s (D) 1609 m/s A. Correct! You need to conert minutes to seconds, one mile into meters, then use = d/t. = 1609m / 40sec = 6.7 m/s B. Incorrect! You forgot to use four minutes instead of one. C. Incorrect! You forgot to change minutes into seconds. D. Incorrect! That s way too fast, faster than the speed of sound. Consider the number of seconds in four minutes. You need to conert minutes to seconds, one mile into meters, then use = d/t. 1 mile = 1609 m 4 minutes = 40 seconds V = 1609 m/ 40 sec = 6.7 m/s The correct answer is (A).
Question No. 3 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 3. A car traels on a slick highway at 5 m/s. When applying the brakes, it can slow at -6 m/s/s. How much time is needed to bring it to a complete halt? Question #03 (A) 1.6 sec (B) 4. sec (C) 9.8 sec (D) 58.8 sec A. Incorrect! Consider the acceleration formula/definition. Acceleration equal change in elocity per change in time. B. Correct! Acceleration equal change in elocity per change in time. Rearranging gies time equals change in elocity diided by acceleration. The change in elocity is 5m/s and the acceleration is -6 m/s/s. Diiding these quantities gie 4. seconds. C. Incorrect! That s the acceleration due to graity. The car isn t falling to earth so we won t use that one here. D. Incorrect! Consider the acceleration formula/definition. Acceleration equal change in elocity per change in time. Gien: i = 5 m/s acceleration = -6 m/s f = 0 m/s (halted) Find: time =? a = Δ/Δt= f - i /Δt t = ( f - i )/a =(0-5 m/s) / -6m/s = 4. s The correct answer is (B).
Question No. 4 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 4. A confused and befuddled hiker walks 3km North, then 1km West, then km South, then 3km West, and finally km North. After all this, they sit down and try to use physics to compute their location. How far hae they moed from their original starting position? Question #04 (A) 5 km (B) 11 km (C) 15 km (D) 45 km A. Correct! Add similar components. Add up all the North/South moements, getting a resultant for that direction. Add up all the East/West moements, getting a resultant for that direction. Finally, combine those two resultants to get his final distance from the starting point. Use the Pythagorean theorem to get the magnitude of the resultant. B. Incorrect! For distances that are opposite the assigned positie must be made negatie. For example, if North is assigned as positie, any South moements must be negatie. C. Incorrect! Don t simply add all the alues, direction must be taken into account. D. Incorrect! Don t forget to take the square root when you are finding the final resultant. Adding all the North/South moements gie +3km North. Adding all the East/West moements gies -4km West. Using these two as the sides of a right triangle, find the hypotenuse with the Pythagorean theorem. c a b c a b c (3km) ( 4km) c 5km 5. 0 km It would be typical to assign North as positie, and East as positie. Howeer, if that were reersed, as long as it was consistently used, the same answer would be found. Also, the problem doesn t ask for a direction, but if it were to be found, trigonometry could be used to find whateer angle was asked. The correct answer is (A).
Question No. 5 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 5. An object is thrown downwards at 0 m/s from the top of the Empire State building in New York. This building is 31m tall. Neglecting the effects of air resistance and friction, find how fast will the object be moing when it gets to the ground below. Question #05 (A) +81.8 m/s (B) +76.7 n/s (C) +0m/s (D) +9.8m/s A. Correct! Notice that the initial elocity and acceleration due to graity are both in the same direction. They must hae the same sign. In this case, they are both made to be +. Use f =i + ad here. The initial elocity is 0 m/s, the distance is 31m, and the acceleration from graity is 9.8 m/s. B. Incorrect! You must account for the initial elocity of 0 m/s downward. C. Incorrect! The object will not fall at a constant elocity due to the acceleration from graity. Use a formula that incorporates that fact. D. Incorrect! 9.8 is the magnitude of the acceleration from graity. Howeer, that isn t the same as the elocity of the object. Identify the information gien, along with the unknown, and look for a formula that incorporates all of those. Gien: i = 0 m/s this elocity will arbitrarily be made +. a = 9.8 m/s. Since we hae made the down direction +, this too is +. d= 31m f =? Use: f i ad f 0m/s 9.8m/s 31m 400m /s 690m /s m f 6690 /s f 6690m /s 81.8m/s The correct answer is (A).
Question No. 6 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 6. A motorcyclist throws a ball into the air ahead of him, always staying under the ball in order to catch it. If the motorcyclist s speed is 5 m/s, and the initial speed of the ball is 50 m/s, how long is the time of flight? Question #06 (A) 4.3 s (B) 5.0 s (C) 7.1 s (D) 8.8 s A. Incorrect. This is roughly half of the correct alue. B. Incorrect. The elocity of the ball is not 5 km/s. C. Incorrect. The ball is not thrown up at a 45 angle. D. Correct! Splitting the moement of the ball into x and y components and calculate the time going upwards. The time of flight is twice as much. First, find both horizontal and ertical components of the ball motion. Since the ball is always under the motorcycle, we know what the x-component of the ball s initial elocity is. Since we hae the horizontal component, and the hypotenuse of 50 m/s, we can use the Pythagorean theorem to find the ertical component. c =a +b 50m/s = 5m/s + V V =43.3m/s Next, use the definition of acceleration. The change in ertical elocity of the ball will be twice the amount we calculated since it goes up, and returns downwards at an equal speed. If t(up) is the time to take going upwards, the initial speed of ertical elocity is Vi = 43.3 m/s and the final speed ertically is V f = 0, we can use the acceleration formula, where V (ertical) = V f - V i wher, to calculate the time taken to go upwards, t(up): Δ a= t Δ 0m/s-43.3m/s t(up) = = = 4.4 s a -9.8m/s/s Hence, t(total) = t(up) + t(down) = x t(up) = 8.8 s (D) is the correct answer.
Question No. 7 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 7. A balloon is at an altitude of 80m. It is moing upwards at a constant 10m/s when a package is dropped. How long does it take for the package to reach the ground? Question #7 (A) -3.1 sec (B) 4.0 sec (C) 5.sec (D) 8.0 sec A. Incorrect! This is a solution for the quadratic equation obtained with the gien information, but it is an unrealistic answer since it s a negatie time. This is not the actual answer. B. Incorrect! This would be the time if there was no initial elocity of the package. Howeer there is so that must be taken into account. C. Correct! Substitute the gien alues into d= i t+at /. The gien distance and acceleration can be taken as negatie since they are both downward. The initial elocity is positie. Sole the quadratic equation for t. D. Incorrect! This would be the time if the package dropped at a constant elocity of 10 m/s. The acceleration due to graity is known. The initial distance is taken as negatie since the package will be falling down. The initial elocity is also the initial elocity of the package. We want to find the time it takes for the package to hit the ground. Use: d=it+at / This will be a quadratic equation. Substitute alues and notice the coefficients for the quadratic formula. The units hae been dropped to clarify the quadratic relationship. -80=10t-9.8t / -4.9t +10t+80=0 This is a quadratic equation in the form of ax + bx + c = 0, where a = -4.9; b = 10; c = 80. Substituting into the quadratic formula gie: -b± b -4ac -10± 10-4(-4.9)80 = a (-4.9) Soling this expression gies two solutions -3.1 and 5.. The negatie alue will be disregarded. The realistic positie answer is t = 5. sec. You can plug that alue back in to double check the equality. Correct answer is (C).
Question No. 8 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 8. A manned motorboat heads west at 4.0 km/h across a south flowing rier which flows at 3.0 km/h. If the manned motorboat traels for 1 min, how far has it gone? Question #08 (A) 0.60 km (B) 0.80 km (C) 1.0 km (D) 1.4 km A. Incorrect. The speed of the boat must be taken into account. B. Incorrect. The speed of the rier current must be taken into account. C. Correct! See below for the correct solution. D. Incorrect. Splitting the moement of the cannonball into components would be a good idea. First, find the net elocity, letting the x-direction be east, and letting the y-direction be north. Then conert the total time in minutes into hours. Then use the equation distance= (elocity) (time). =-4.0 km/h x =-3.0 km/h y = + = (-4.0) +(-3.0) =5.0 km/h x y 1 h 1 Δt= 1 min = h 60 min 5.0 1 d=δt= 5.0 km/h h =1.00 km 5.0 The correct answer is (C).
Question No. 9 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 9. A ball is dropped. Which of the plots below represent the elocity as a function of displacement? (A) Question #09 (B) (C) (D) A. Correct! =ad, so d. We are looking for a square root relationship, which is what the graph represents. B. Incorrect. The relationship between elocity and displacement is not linear. C. Incorrect. This graph assumes d. D. Incorrect. The elocity is not constant. We are looking for a square root relationship, which is what the graph represents. Recall that V f =V i +ad. Since elocity depends upon the square root of d, the graph would look like selection A. The correct answer is (A).
Question No. 10 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 3 10. Suppose that the position of an object as a function of time is x(t)=t -t+1, where t is in seconds and x is in meters. What is the aerage elocity of the object between t = 1.0 s and t =.0 s? Question #05 (A) 1.0 m/s (B) 5.8 m/s (C) 6.0 m/s (D) 6.5 m/s A. Incorrect. Finding the displacement oer the time interal would be a good idea. B. Incorrect. You cannot just ealuate the deriatie at 1.5 s. C. Correct! Use the formula that aerage elocity is the displacement diided by the time interal, as shown in the solution below. D. Incorrect. You cannot take the deriatie at 1.0 s and.0 s and aerage. Δx x()-x(1) = = Δt -1 x(1)=1 x()=7 =6 m/s Answer (C) is correct.