Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

A particlemoves in thexy plane withconstant acceleration. At time zero, theparti

ID: 1735426 • Letter: A

Question

A particlemoves in thexy plane withconstant

acceleration. At time zero, theparticle is at

x =4.5 m,y =8.5 m, and hasvelocity ~vo =

(5.5 m/s) ˆi + (8 m/s) ˆ? . The accelerationis

given by ~a = (2 m/s2) ˆi + (5.5 m/s2) ˆ?.

What is the x component of velocityafter

3.5 s? Answer in units ofm/s.

(part 2 of 3)

What is the y component of velocityafter

3.5 s? Answer in units ofm/s.

part 3 of 3

What is the magnitude of thedisplacement

from the origin(x = 0m, y = 0 m)after

3.5 s? Answer in units ofm

QuestionDetails:

A particlemoves in thexy plane withconstant

acceleration. At time zero, theparticle is at

x =4.5 m,y =8.5 m, and hasvelocity ~vo =

(5.5 m/s) ˆi + (8 m/s) ˆ? . The accelerationis

given by ~a = (2 m/s2) ˆi + (5.5 m/s2) ˆ?.

What is the x component of velocityafter

3.5 s? Answer in units ofm/s.

(part 2 of 3)

What is the y component of velocityafter

3.5 s? Answer in units ofm/s.

part 3 of 3

What is the magnitude of thedisplacement

from the origin(x = 0m, y = 0 m)after

3.5 s? Answer in units ofm

Explanation / Answer

X direction velocity=5.5 accleration=2 velocity after 3.5 sec=v v=u+a*t v=5.5+2*3.5 v=12.5 m/s (ans part 1) Y direction velocity=-8 accleration=5.5 the velocity and accleration are opposite direction.so the particalhas negative accleration at first let the velocity=0 after t sec v=u-a*t 0=8-5.5*t t=1.45 sec in 3.5-1.45=2.045 sec the veloity =v1 v1=u+a*t v1=0+5.5*2.045=11.2475 m/s (ans part 2) X direction dispcement in 3.5 sec=s s=u*t+(1/2)*a*t2 s=5.5*3.5+(1/2)*2*3.5*3.5 s=31.5 Y direction in the first 1.45 sec the partical move towards negative Y axis letthat displacement=s1 s1=u*t-(1/2)*a*t2 s1=8*1.45-(1/2)*5.5*1.45*1.45 s1=5.81 m in the next 2.045 sec the partical move s2 distance in the positiveY direction s2=u*t+(1/2)*a*t2 s2=0+(1/2)*5.5*2.045*2.045 s2=11.5 the net displacement in 3.5 sec=11.5-5.81 m=5.69 m after 3.5 sec the partical is at position X=31.5+4.5=36 m and Y=14.19 m (ans part 3)

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote