PHYS2091_U2 AS(1)

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Kean University *

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2091

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Mathematics

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Apr 25, 2024

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docx

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PHYS2091 Unit 2 AS: Homework 2 Description: The purpose of this assignment is to develop problem-solving skills using real-life examples. The students will engage in problems ranging from everyday activities to the applications of physics in biology. Instructions: This assignment has 15 questions of various complexities. Upload your answers as a single PDF file on Blackboard. 35 points CSLO 1, 2, 3 1. (2 points) A commuter backs her car out of her garage with an acceleration of 1.40 m/s 2 . (a) How long does it take her to reach a speed of 2.00 m/s? 1.40= 2.00-0/t T= 2.00/1.40= 1.43s (b) If she then brakes to a stop in 0.800 s, what is her deceleration? 0-2.00/0.800= -2.00/0.800= -2.50 m/s 2 2. (2 points) An Olympic-class sprinter starts a race with an acceleration of 4.50 m/s 2 . (a) What is her speed 2.40 s later? 4.50= final speed/2.40= 10.8m/s (b) Sketch a graph of her position vs. time for this period. 3. (3 points) Blood is accelerated from rest to 30.0 cm/s in a distance of 1.80 cm by the left ventricle of the heart.

(a) List the knowns in this problem. Initial velocity is 0, distance is 1.80cm, and final velocity is 30.0cm/s. (b) How long does the acceleration take? To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking your units. Change in x= ½(Vi+Vf)t because we have x, vi, and vf, we need to find time. 1.80= ½(0+30.0)t 1.80=15.0t T= 0.12s 4. (2 points) A powerful motorcycle can accelerate from rest to 26.8 m/s (100 km/h) in only 3.90 s. (a) What is its average acceleration? Vf= Vi+at 26.8= 0+a(3.90) 26.8= 3.90a A= 6.87 m/s 2 (b) How far does it travel in that time? Change in x= ½(Vi+Vf)t =½(0+26.8)(3.90) =1/2(26.8)(3.90)= 52.3m 5. (4 points) For a ball thrown straight up with an initial velocity of 15.0 m/s, where the point of release is chosen to be y o = 0 m, calculate the displacement and velocity at times of (a) 0.500 s, Vf= 15+(-9.80)(0.500)= 10.1m/s X= 15.0(0.500)+1/2(-9.8)(0.500)^2 =7.50+(-1.23)= 6.23m (b) 1.00 s, Vf= 15+(-9.80)(1.00)= 5.2m/s X= 15.0(1.00)+1/2(-9.8)(1.00)^2 =15-4.90= 10.1m (c) 1.50 s, and Vf= 15+(-9.80)(1.50)= 0.30m/s X= 15.0(1.50)+1/2(-9.8)(1.50)^2 =11.475m (d) 2.00 s. Vf= 15+(-9.80)(2.00)= -4.6m/s X= 15.0(2.00)+1/2(-9.8)(2.00)^2 =10.4m

6. (5 points) The Verrazano Narrows Bridge in New York City is 70.0 m above the water. If a rock thrown straight down with an initial velocity of 14.0 m/s from the Bridge, calculate the displacement and velocity at times of (a) 0.500 s, Vf= vi+at = -14.0+(-9.8)(0.500) = -18.9m/s X= -14.0(0.500)+1/2(-9.8)(0.500)^2 = -7.00-1.23 = -8.23m (b) 1.00 s, Vf= vi+at = -14.0+(-9.8)(1) = -23.8m/s X= -14.0(1.00)+1/2(-9.8)(1.00)^2 = -14.0-4.90 = -18.9m (c) 1.50 s, Vf= vi+at = -14.0+(-9.8)(1.50) = -28.7m/s X= -14.0(1.50)+1/2(-9.8)(1.50)^2 = -32.025m (d) 2.00 s, and Vf= vi+at = -14.0+(-9.8)(2.00) = -33.6m/s X= -14.0(2.00)+1/2(-9.8)(2.00)^2 = -47.6m (e) 2.50 s Vf= vi+at = -14.0+(-9.8)(2.50) = -38.5m/s X= -14.0(2.50)+1/2(-9.8)(2.50)^2 = -65.625m 7. (2 points) A swan on a lake gets airborne by flapping its wings and running on top of the water. (a) If the swan must reach a velocity of 6.00 m/s to take off and it accelerates from rest at an average rate of 0.350 m/s 2 , how far will it travel before becoming airborne? 6.00^2= 0^2+20.350)(x) 36.0= 0.700x X= 51.4m (b) How long does this take? 6.00= 0+(0.350)t

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