233012125-Dyna-Project.docx - EXAMPLES/NOTES UNIFORM...

This preview shows page 1 - 10 out of 51 pages.

EXAMPLES/NOTES: UNIFORM RECTILINEAR MOTION 1. Car A at a gasoline station stays there for 10 minutes after Car B passes at a constant speed of 40 miles per hour. How long will it take Car A to overtake Car B if it accelerates at 4 m/s² Solution: Ta = Tb Sa = S10 + Sb Consider B: v = s t Tb = s v = Sb 17.88 m / s Sb = 17.88 Tb Consider A: Sa = VoTa + 1 2 aT ² = 0 + 1 2 ( 4 m s 2 ) ( T a 2 ) = 2 Ta ² For S10: S 10 = vt = ( 17.88 m s ) ( 600 s ) = 10728 m Substitute: Sa = S 10 + Sb 2 T ² = 10728 + 17.88 T
Answer: T = 77.85 seconds 2. 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 1 s = Vot 1 2 > ² 400 ft = ( 300 ft s ) ( t ) 1 2 ( 32.2 ft s 2 ) ( t 2 ) t = 17.19 s , 1.45 s Consider 2 s = Vot + 1 2 gt 2 200 = 0 + 1 2 ( 32.2 ft s 2 ) ( t 2 ) t = ± 3.52 s How long must one wait? t = 17.19 seconds – 3.52 seconds Answer: 13.67 seconds VARIABLE ACCELERATION 1. The motion of a particle is given by s = 2 t 4 t 3 6 + 2 t 2 where s is in feet and t in seconds . Compute the values of v and a when t = 2 seconds .
s = 2 t 4 t 3 6 + 2 t 2 v = 8 t 3 1 2 t 2 + 4 t a = 24 t 2 t + 4 @t = 2 seconds v = 8 ( 2 ) 3 1 2 ( 2 ) 2 + 4 ( 2 ) ; v = 70 ft sec a = 24 ( 2 ) 2 ( 2 ) + 4 ; a = 98 ft sec 2 2. A particle moves in a straight line according to the law s = t 3 40 t where s is in m and t in seconds . a. When t = 5 seconds , compute v . b. Find the average velocity during the 3 rd to 4 th seconds. c. When the particle comes to stop, what is its acceleration? a. s = t 3 40 t v = 3 t 2 40 a = 6 t @t = 5 seconds v = 3 ( 5 ) 2 40 ; v = 35 m sec
b. v ave = total distance totaltime = 96 −(− 93 ) 4 3 = 3 1 ; v ave =− 3 m sec @ 3 rd second s = ( 3 ) 3 40 ( 3 ) ; s =− 93 m @ 4 th second s =( 4 ) 3 40 ( 4 ) ; s =− 96 m c. if v = 0 0 = 3 t 2 40 t = 3.65 seconds a = 6 t = 6 ( 3.65 ) ; a = 21.9 m sec 2 3. The rectilinear motion of a given particle is given by s = v 2 9 where s is in m and v is in m sec . When t = 0 , s = 0 , and v = 3 m sec . Determine s t , v t , and a t relations. s = v 2 9 ds dt = 2 v dv dt v = 2 va a = 1 2 a = v v o t
1 2 = v −( 3 ) t ; v = 1 2 t + 3 s = v o t + 1 2 at 2 = ( 3 ) t + 1 2 ( 1 2 ) t 2 ; s = 3 t + 1 4 t 2 MOTION CURVES 1. A particle starting with an initial velocity of 60 ft /s has a rectilinear motion with the constant deceleration of 10 ft/ s 2 . 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/ s 2 , starting from rest at A and coming to stop at B. Find the maximum speed in fps. Solution:
120 ) V A 6x – 6(40 – x) = 0 6x – 0 = 6(40 – x) 6x = 240 – 6x x = 20 3. 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.

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture