Question

Chapter 2, Practice Problem 2/015 A particle starts from rest at x = -3.7 m and moves along the x-axis with the velocity history shown. Plot the corresponding acceleration and the displacement histories for the 4.4 seconds. Find the time t when the particle crosses the origin. After you have the plots, answer the questions. , m/s 5.1 4,4 2.2 1,7 Questions At t 0.73 s m/s2 m/s, a At t 1.41 5, m/s, a m/s2 m v At t 2.74 5,¥ = m/s, a m/s2 m, v = The particle will cross the origin when t=

Answer 1

acceleration of is the rabe of change of posiion it's parbele and Velocity of change o Yelociny ot parhdle rate he is the inibial po si tion of the Parkele -3.7 m. Given that is Of the paykcle. displace ment The under the graph V-L is are a The slope of the graph of v-t is the Y (m/sa) pavkide he B a ccelerahon A 5.1 to t 1.1 sec, t, (0) From = 0 Sec displace ment S area of e o AB E C 4.4 3.3 -1 x 5.1= +2.805 m 2.2 2 So, he Posihon ti at is (3.7) +(2.805) 0.895 m. O A straight line and has is because is Constant acceleration 1he 4.64 mse S.1 slope Con stant a= So .

the velocity 5.1 m/sec is Gonstant V = From t,=1 to t 2.2 sec, m/se accelevabion a 2 So of rectangle ABGF The diplacament S2 ave a is (5.)(2 .2 - 1-1) = (5.)x (1.1) 5.6 m. S, = +4.715 m. -0.895 + 5.61 Posi hon at So, he is t, displace ment to he is From t, 3.85 Sec, 2.2 sec +4.2075 m Area oAe 5.1 x .65 BG C 4.715 + 4.2075 is So, the posi tion at = + 3 t, =om /sec At +8.9225 m. 0 S. The acceleration duning - 3.09 mse +his period will a = be 3.85-2-2 (iv) So, the velocity at a, (tts) 4.4 Sec will V3 be -L.7 m/sec 0 + 309) 0.5s) displacement during So how the penod S = 7)0.55) this iS = -0.4675m Posi bon at the i s So 3+S 4.2075-0.4675 = + 3.74 Now let's Plot graphs for posi bion (z-t) accelerabon (a-t) and a 7.5 4.5 4-64 1-5 2.2 3.3 4.4 3.3 4.4 2.2 -3.09

(a) displacament fom t At t o-73 Sec, = 0 Sec iS (-3.741.236) Posi hon is a -.236 m i4.64 - 0 +(4.64)(0.7) a, Velocty m) ス。= -2.464 Posi hon So, the is 4.64 m/se acceleyahon a, a is 3.3872 m/sec Ya = V, (t- ) is dis placement trem t 11 sec (b). At =1-41 Sec, position , +5ib -0.895 1-58 So is = 1.581 m, Sh(5.1)(0. 31) m/sec a m/sec, abo = 5.1 0 .686 m is S + t,2.2 s ec (c) At te2.74 sec, displacement trom 12 (3.04) (0. S4) E 4-65 m = 5.1 5.1 m/sec, S2c VV= 23 tet) t9.365 m, Ve= う。= + Se 4.715 4.65 -3.09 m/sec e =t3.4314 m/sec, a - a Ve= 5.1 -3.09)(o.Su) So the answer S ave now a 4-64 m2 3.38 72 m V At to.73s, - 2.464 m Oms ら.1 V = t 1-4l S, 0.686 m At - 3.09 ms a = Y 3.4 314 me At +9.365 m t 2.74 S, point t the parbicle CYo sser ongin ie xp o Now lel's say at Some belween t, and t is See that From (-t) g aph cas w e (-0.895) +0.895 S1p 0- , S Ip 0.895 but (s.)(tp-1) (5.1)(tp- 1.1) but S1p s -2755 Sec (tr-1.1) =0.1755 when 2755 s onigin Cross Hhe So, the pavhcle will