EXAMPLES/NOTES:UNIFORM RECTILINEAR MOTION1. Car A at a gasoline station stays there for 10 minutes after Car B passes at a constant speed of 40 milesper hour. How long will it take Car A to overtake Car B if it accelerates at 4 m/s²Solution:Ta = TbSa = S10 + SbConsider B:v=stTb=sv=Sb17.88m/sSb=17.88TbConsider A:Sa=VoTa+12aT²=0+12(4ms2)(T a2)=2Ta²For S10:S10=vt=(17.88ms)(600s)=10728mSubstitute:Sa=S10+Sb2T²=10728+17.88T
Answer: T = 77.85 seconds2. A stone is thrown up from the ground with a velocity of 300ft/s. How long must one wait before dropping a second stone from the top of 600ft tower if the two stones are to pass each other 200ft from the top of the tower?Solution:Consider 1s=Vot−12>²400ft=(300fts)(t)−12(32.2fts2)(t2)t=17.19s ,1.45sConsider 2s=Vot+12gt2200=0+12(32.2fts2)(t2)t=±3.52sHow long must one wait?t = 17.19 seconds – 3.52 secondsAnswer: 13.67 secondsVARIABLE ACCELERATION1. The motion of a particle is given by s=2t4−t36+2t2where sis in feetand tinseconds. Compute the values of vand awhent=2seconds.
s=2t4−t36+2t2v=8t3−12t2+4ta=24t2−t+4@t=2secondsv=8(2)3−12(2)2+4(2); v=70ftseca=24(2)2−(2)+4; a=98ftsec22. A particle moves in a straight line according to the law s=t3−40twhere sis in mandtin seconds.a.When t=5seconds, compute v.b.Find the average velocity during the 3rdto 4thseconds.c.When the particle comes to stop, what is its acceleration?a.s=t3−40tv=3t2−40a=6t@t=5secondsv=3(5)2−40; v=35msec
b.vave=total distancetotaltime=−96−(−93)4−3=−31; vave=−3msec@3rdseconds=(3)3−40(3); s=−93m@4thseconds=(4)3−40(4); s=−96mc.if v=00=3t2−40t=3.65secondsa=6t=6(3.65); a=21.9msec23. The rectilinear motion of a given particle is given by s=v2−9where sis in mand vis in msec. When t=0, s=0, and v=3msec. Determine s−t, v−t, and a−trelations.s=v2−9dsdt=2vdvdtv=2vaa=12a=v−vot
12=v−(3)t;v=12t+3s=vot+12at2=(3)t+12(12)t2;s=3t+14t2MOTION CURVES1. A particle starting with an initial velocity of 60 ft /s has a rectilinear motion with the constant deceleration of 10 ft/s2. Determine the velocity and displacement at t = 9 sec.Solution:
For and :2. An auto travelled 1800 ft in 40sec. The auto accelerates uniformly and decelerates uniformly at 6 ft/s2, starting from rest at A and coming to stop at B. Find the maximum speed in fps.Solution:
120)VA6x – 6(40 – x) = 06x – 0 = 6(40 – x)6x = 240 – 6xx = 203. An Auto starts from rest and reaches a speed of 60 ft/s in 15 sec. The acceleration increases uniformly from zero for the first 9 sec after which the acceleration reduces uniformly to zero in the next 6 sec. Compute for the displacement in this 15 sec interval.Solution:
V:For and :
PROJECTILE MOTION 1. A golf ball is fired from the top of a cliff 50 m high with a velocity of 10 m/s directed at 45⁰ to the horizontal. Find the range of the projectile.