(5%) Problem 23: A detailed graph of acceleration versus time is shown. am/s +5.0 5.0 10.0 15.0 20.0 25.0 t(s) 5.0 20% Part (a) What is the instantaneous acceleration at time 17.75 s? a= - 8 -8.000 m/s / Correct! 20% Part (b) What is the change in velocity during the time interval from 1.25 s to 8. 75 s? Av = 37.50 m/s / Correct! * 20% Part (c) What is the change in velocity during the time interval from 8. 75's to 17. 75 s? Av =1 m/s sin( cos() tan() z C 7 8 9 cotan() asino acos() 4 5 6 atan() acotan() sinh( 1 2 3 cosh( tanho cotanh() ODegrees @ Radians ted Submit Hint I give up! Hints: 0% deduction per hint Hints remaining. Feedback: An deduction per feedback. ted 20% Part (d) If the initial velocity is / m/s. then what is the velocity at time (9.S s? V- - 32 Feedback: is available. v = -32.00 m/s X. Attempts Rezanin 20% Part (e) What is the average acceleration during the time interval from 8. 35's to 245 s? a- -2-71 a- 2.710 m/s Correct!The Expert TA | Human-like Grad X Do Homework - Section 1.4 X Ita The Expert TA | Human-lik tta.com/Common/TakeTutorialAssignment.aspx Class Management Help Week3 Begin Date: 6/15/2023 11:59:00 PM -- Due Date: 6/27/2023 11:59:00 PM End Date: 8/10/2023 11:59:00 PM (6%) Problem 1: A golfer hits two consecutive shots in which the ball's displacements are Ari - (301 m) i and Ar2 - (187 m) i + (183 m) j 50% Part (a) What is the magnitude, in meters, of the net displacement? d = m sin( Cos() can( 7 8 9 cotan() asin() acos() E 4 5 6 atan( acotan() sinh( 2 3 cosh( "tanh() cotanh() 0 END Degrees O. Radians VO Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: ! Feedback: 0% deduction per feedback. 50% Part (b) What angle, in degrees, does the net displacement vector make with the positive y axis? All content @ 2023 Expert TA, LLC Q SearchX ta The Expert TA | Human-like Grad X P Do Homework - Section 1.4 x ta The Expert TA | Huma pertta.com/Common/TakeTutorialAssignment.aspx Week3 Begin Date: 6/15/2023 11:59:00 PM -- Due Date: 6/27/2023 11:59:00 PM End Date: 8/10/2023 11:59:00 PM (6%) Problem 2: The position of a particle is given by the following expression: r(t) - (3.74 m/s? ) t? i + (5.53 m) 3 + [(3.02 m/s) t] k 33% Part (a) What is the magnitude, in meters per second, of the velocity of the particle at t - 2.95 s? v(2.95 s) = m/s sin( cos() tan( 7 8 19 cotan( asin( acos() E 4 5 6 atan() acotan() sinh() 2 3 cosh( tanh() cotanh() + 0 Degrees O Radians VO KSPACE CLEAR Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: 3 Feedback: 0% deduction per feedback. & 33% Part (b) What angle, in degrees, does the velocity of the particle make with the x axis at t = 2.95 s? 6 33% Part (c) What is the magnitude, in meters per squared second, of the particle's acceleration? All content @ 2023 Expert TA, LLC Q Search O V O CX ta The Expert TA | Human-like Grad x Do Homework - Section 1.4 x ta The Expert TA | Human- pertta.com/Common/TakeTutorialAssignment.aspx Class Management | Help Week3 Begin Date: 6/15/2023 11:59:00 PM -- Due Date: 6/27/2023 11:59:00 PM End Date: 8/10/2023 11:59:00 PM (5%) Problem 3: A particle's initial position is To - (3.36 m) i + (3.67 m) j It undergoes a displacement given by Ar - (2.13 m) i + (2.59 m) j + ( - 5.53 m) k What is the magnitude, in meters, of the particle's final position? m ed ed sin() cos() tan() 7 8 9 HOME cotan() asin() acos() 4 5 6 atan() acotan() sinh() 2 3 cosh() tanh() cotanh() O END O Degrees O Radians CLEAR Hint ted Submit Feedback I give up! Hints: 0% deduction per hint. Hints remaining: 2 Feedback: 0% deduction per feedback All content @ 2023 Expert TA, LLC Q Search Ota The Expert TA | Human-like Grad X P Do Homework - Section 1.4 x ta The Expert TA | Human-like Grad X a.com/Common/TakeTutorialAssignment.aspx + Q Week3 Begin Date: 6/15/2023 11:59:00 PM -- Due Date: 6/27/2023 11:59:00 PM End Date: 8/10/2023 11:59:00 PM (6%) Problem 4: A car moves along a horizontal road with constant velocity UO = VO,I i U 1 = (33.6 m/s) i until it encounters a smooth inclined hill. It climbs the hill with constant velocity U1 = V1,z i + V1,y j = (34.2 m/s) i + (6.23 m/s) j as indicated in the figure. The period of time during which the car changes its velocity is At = 3.19 s. 50% Part (a) Enter an expression, in Cartesian unit-vector notation, for the average acceleration of the car during the given time period using the symbols provided. davg = I Grade Summary Deductions 0% Potential 100% At ( 7 8 9 HOME Submissions 4 5 6 Attempts remaining; 999 d g (0% per attempt) K m vo 1 2 3 detailed view VOX V 1 V 1x 0 Vly VOl Submit Hint I give up! Feedback: 0% deduction per feedback. Hints: 0% deduction per hint. Hints remaining: 1 4 50% Part (b) What is the magnitude, in meters per squared second, of the car's average acceleration during the given time period. All content @ 2023 Expert TA, LLC Q Search 5ta The Expert TA | Human-like Grad X P Do Homework - Section 1.4 a.com/Common/TakeTutorialAssignment.aspx x ta The Expert TA | Human-like Grad X + Q 1 X weeks Begin Date: 6/13/2023 11:59:00 PM -- Due Date: 6/27/2023 11:by:00 PM End Date: 8/10/2023 11:39:00 PM (10%) Problem 5: While competing in the long jump, a person leaps over a smooth horizontal sand surface. She lands on the surface with speed ve = 9.75 m/s at an angle 0 = 30.50 below horizontal. Assume that the person moves without air resistance. Use a Cartesian coordinate system with the origin at her final position. The positive x-axis is directed from her initial to her final position, and the positive y-axis is directed vertically upwards. 33% Part (a) Enter an expression for the jumper's initial velocity vector in terms of vf, 0, g, and the unit vectors i and j. Vo = Grade Summary Deductions Potential 100% cos(a) cos( Q) cos(0) ( ) 7 8 9 Submissions sin(a sin(Q) sin(0) 4 5 6 Attempts remaining: 999 d 1 2 3 0% per attempt) detailed view + 0 m VO Submit Hint I give up ! Hints: 0% deduction per hint. Hints remaining: Feedback: 0% deduction per feedback. 4 33% Part (b) Calculate the maximum height, in meters, that the jumper reaches above ground level. A 33% Part (c) What x position, in meters, did the jumper begin her long jump? All content @ 2023 Expert TA , LLC Q Search 6/ 2 - FB 8 3 E R G