(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!(6%) Problem 24: A detailed graph of velocity versus time is shown. 30.0 20.0 v(m/ s 10.0 5.0 10.0 15.0 20.0 25.0 t(S) ted 17% Part (a) What is the velocity at time 10.25 s? = 30.00 m/s / Correct! ted 17% Part (b) What is the acceleration at time 10.25 s? ated a= 0.000 m/s Correct! eted ted 17% Part (c) What is the acceleration at time 17 s? a= - 2 leted -2.000 m/s / Correct! leted pleted *17% Part (d) What is the displacement during the time interval from 3.5 s to 10.25 s? pleted Ax - pleted m apleted apleted sino cos tano 7 8 9 npleted cotan asinO acos( E 4 5 6 tial atano acotan( sinho 1 2 3 rial cosho tanho cotanho O ntial Degrees @ Radians IVO ompleted Submit Hint I give up! artial Completed Hints: 0% deduction per hint Hints remaining: Feedback: OPu deduction per feedback Partial Partial 17% Part (e) If the initial position is -25 m. then what is the position at time . s? x = 425.0m / Correct! $ 17% Part (1) What is the average velocity during the time interval (0. is to 2 75 s? V - 425.0 m/s X Attempts RemainClass Management | Help theExpertTA.com 1 Stu Week2 Begin Date: 6/9/2023 12:01:00 AM -- Due Date: 6/20/2023 11:59:00 PM End Date: 8/10/2023 11:59:00 PM coming to rest in to = 5.63 s. (4%) Problem 25: Referring to the figure, a jeep travels a distance d - 26.6 m in the positive r direction in a time to = 17.2 s, at which point the jeep brakes, y1 ed 25% Part (a) What was the jeep's average velocity in the horizontal direction, in meters per second, during the to time period? ed Warg = 1.547 m/s / Correct! ted ted interval ti? * 25% Part (b) Assuming the jeep started from rest and moved with a constant acceleration, what was the acceleration, in meters per squared second, during time ted ated a = m / $2 Grade Summary Deductions 09% Potential 1009% leted sin() cos() tan( TI ( ) 7 8 9 Late Work 96 100% Late Potential 100% leted cotan() asin() acos() E 4 5 6 leted atan() acotan() Sinh() 1 2 3 Submissions pleted Attempts remaining 994 cosho tanh() cotanh() (09% per attempt) pleted Degrees O Radians CLEAR detailed view pleted pleted 109% Submit Hint Feedback I give up! pleted nplete Hints: 09% deduction per hint Hints remaining Feedback: 0% deduction per feedback. mpleted * 25% Part (c) What was the jeep's instantaneous velocity in the horizontal direction, in meters per second, when it began braking? mpleted v1 =-3.2604 mpleted v1 = -3.260 m/s X Attempts Remain ntial intial * 25% Part (d) Using the result from Part (c), what was the jeep's horizontal component of acceleration, in meters per squared second, during the braking period? artial at - 0.5790 m/s? X Attempts Remain ompleted artial Partial Completed Partial Partial &Class Management | Help Week2 Begin Date: 6/9/2023 12:01:00 AM -- Due Date: 6/20/2023 11:59:00 PM End Date: 8/10/2023 11:59:00 PM (2%) Problem 27: While entering a freeway, a car accelerates from rest at a rate of 2.9 m/s for 14.5 s. 50% Part (a) How far does the car travel in those 14.5 s, in meters? d = 304.86 d = 304.9 Correct! $ 50% Part (b) What is the car's final velocity, in meters per second? sing cos tan( 7 8 9 cotan () asin() acos() E 4 5 6 atan() acotan() sinh() 1 2 3 cosho tanh() cotanh() Degrees Radians VO Submit Hint I give up! ed ted Hints: 0) deduction per hint. Hints remaining Feedback: 0deduction per feedback. ted ted eted eted leted pleted alClass Management Help theExpertTA. Week2 Begin Date: 6/9/2023 12:01:00 AM -- Due Date: 6/20/2023 11:59:00 PM End Date: 8/10/2023 11:59:00 PM (4%) Problem 28: A particle moves in a straight line with a constant acceleration of 5. 72 m/s in the positive x direction. 5 50% Part (a) If its initial velocity is 1.65 m/s in the positive x direction, then how long, in seconds, does it take to move 5.1/ m? - 3.460 X Attempts Remain $ 50% Part (b) What is the particle's final velocity in m/s? Grade Sum Deductions Potential sin() cos() TE ( ) 7 8 9 Late Work % tan() HOME Late Potential cotan() asin( acos() E 4 atan() acotan() sinh() 1 2 3 Submissions Attempts remai cosh( tanh() cotanh() END (09% per attempt Degrees O Radians VO detailed view CLEAR Submit Hint I give up! Hints: 0% deduction per hint, Hints remaining: Feedback: 0% deduction per feedback hs All content @ 2023 Expert 1A, LLCI am sorry there was a miscalculation in my part. You need to replace 5 with -8 in the acceleration value between 148 to 198. Following that the value for change in velocity becomes 23.75m/s (Part 0 solution). 16m/s is the solution for Part d. Just calculate the change in velocity in between Os and 19.58. Then add the change in velocity which comes to 9m/s to initial velocity 7m/s. The final answer comes to be 16m/s. I am sorry for the mistake. I apologise