My Account lomewor 02 Begin Date: & 00 00 PM-Due Date PM End Date: 9 (11%) Probl

Question

My Account lomewor 02 Begin Date: & 00 00 PM-Due Date PM End Date: 9 (11%) Problem 7: A yeep travels a distane d-20. S--the positive x-direction time of h 23& The jeep immediately brakes and comes to rest in t 8 a Randemized Variable d 20.5 n=23 s ent Status here for Otheespertta.com Stata -q) 25% Part (a) what was the jeep's average velocity m the horrental direct on, m meters per second, damen? CompletedC Completed Completed 1 1-25% Part (b) what was the acceleration, in meters p r second squared, daratne interval n. ssurine the jeep zarted fiom restand moved Completed with a constant acceleration Completed0039 Completed 25% Part (c) what was de jeep sttiliitus velocity in the horm eral dnectin ntren per iect d, when began braking? Partial . 25% Part (d) Using the rent hom part (e), what was the jeep's honattal aempkeit areal tak in meta, pa moniquant, dene a braking prriod? Grade Summary 6 8

Explanation / Answer

(a) Average velocity during t1 = 20.5/23 = 0.8913 m/s

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(b) we know that

( vi + vf ) / 2 = average velocity

as vi = 0

vf / 2 = average velocity

vf / 2 = 0.891 m/s

vf = 1.7826 m/s

now, using kinematics,

vf = vi + at

a = vf / t

a = 1.7826 m/s / 23

a = 0.0775 m/s2

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(c) Instantaneous velocity at the time of braking = vf = 1.7826 m/s

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(d) Now, as the car stops, vf = 0, vi = 1.7826 m/s

Therefore,

vf = vi + at

0 = 1.7826 m/s + a(8)

a = - 0.222825 m/s2

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