A particle moves in the xy plane with constant acceleration. At time zero, the p

Question

A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 3 m, y = 8.5 m, and has velocity ~vo = (8 m/s) ˆ+ (8 m/s) ˆ . The acceleration is given by ~a = (8.5 m/s 2 ) ˆ + (7 m/s 2 ) ˆ .

A - What is the x component of velocity after 4 s? Answer in units of m/s = 42

B-What is the y component of velocity after 4 s? Answer in units of m/s. 20

C-What is the magnitude of the displacement from the origin (x = 0 m, y = 0 m) after 4 s? Answer in units of m/???

I have solved A,B but couldn't figure C someone help please

Explanation / Answer

part C) In the kinematics

the general equation for position is given by,

r=r0+v0*t +0.5*a*t^2

Given ,

r0( 3,8.5) , a(8.5,7) , vo( 8,-8) and t = 4 s and v4( 42,20)(found in part A) and partB))
So,

r=(3,8.5) +(8,-8)*4+0.5*(8.5,7)*4^2
= (3,8.5) +(32,-32) +(68,56)
= (3+32+68, 8.5-32+56)

= ( 103, 32.5)

So,
rx=0.103 km, ry=0.0325 km

Magnitude = rx^2 + ry^2

= .103^2 + .0325^2

= 0.108 km

= 108 m

Therefore magnitude of displacement = 108 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.