Question

Problem # 4 (Graded) The motion of a particle is defined as x t2-8t7 and y 0.5t2 +2t 4 where x and y are in meters and t is in seconds. Determine the following: (a) The magnitude of the smallest velocity reached by the particle (b) The time, position, and direction of that velocity.

Answer 1

to solve the given question we use following steps

1. calculate velocity vector

2. minimise the magnitude of velocity as shown.

3. then calculate the required data:

solution:

Giren dtas 2. e in metet =tgt+7 t in secomd - 4 To dind ) Magnitude a he smallest elo cy hed by tre paticle. & direction d hat velocrty ) Time, posihi o Soluctrion doD dthe Giveh data is clemly seein that patele moing In 2-D plane theulere In -diechon Velocity at- 8 elbcny rectan- - dy dt Natat fom, veley h hwing ayain verton Can be written as Cat-8)+12) da Myntde velo y 1V+

Az t- +2) Jor minimam Jal matde CA), ePuatan ith mapet oime we iuen dA do ininum Ualue) _ A O et-) Rltt ther dt 2 Qt-8 + (t+2) W 8t- 322t + t-)+2) 8t 32+ t +4-o dR,8 See cheek lot 28) (1otr8)-)+ 2 (adt-8) t to dt2 Pt-8)+(+e)2 +at-)+) (2t-8j* rf)'a 2 2 (tot-48)

2 2 + both Contam 9nare tem of 312 2 So promd Heme om that tz 8.8 A ut minumum +2)j t-8)2 Then Smallest rtude b elociy 2 I n8-8)(2+8+2) -S.366 m|s veocity &) ime do thus To meach tus velouty , Paembau wp T-tdP.8 Seegnd Position direction -8t+7 movement in Sardinate -7- Se m sten dirction monemert in st +2t-4 5-52 m

Theni t1-56,59 and ad t=0 seedhd S (9)- - 4 curve Then dinal pasttion th maspect to Ontial Pasition 9s 2. Patiom7.s(5.52+4) scale Not Cn Pastion7.396 m Puimd Anse (4) direction Uy tanD dlat tomo 4.8 Aachc (feom ve -a 9 -63.43S axts, in qrdloce se ve = 180-63-4Ss 16 S6S C3. 55.9 Rquired Anuwer 9=63.435