Physics Problem Set: Kinematics and Dynamics Solutions

PJKM T- GARCH BSABE 2-1 PROBLEM SET 1 Given: V32 Omls 2. Given distance 190m time. 4.85 VC = 125mls Solution A = 0.02V2 de Ac 56-10t d. 11.320 Solution v=at v=4.8a a= 50- 10t 190.4.9a x48 +11.53 a dv at = 60-1 lot 9=.5.44 m/52 d= 11.52 a d= R-52 (D.49) du= de 63.63 m 0 V= bot -st 125 50+-54 3. Given : speed 3.2mk . a= 0.17an/p distance 40.2 t2- (1) +2500 Solution: to 55 a.speed L Dislance time a.dx -50k-t= at D1=VI I O2= 5.2 + The distance haveled by bus in t time 6 D2 Vitti/2at2 02=1/2 (0.17at2) 1280 =417m Dr 0.08ast a --0.02}2 10 the distance traveled by student --0-02V 125 1-15-00-02 10 = =/t=4.65 5.24 -0.08957'+ 40.2 m 0-08952-5.2t + 40.2-0 b du dk at dr -0-6-021 dx 2(0.0895) 32-0020 125 "I dv = 0.02 in dx E=9.18 1h10 =0 02X 125 x = 126,29m CS Scanned with CamScanner

Distance traveled by the student in time t 4. Given :U : 300 it ls , 0 : 45 * D1: st solution Initial speed D. . 5.2 x 9.18 D. = 49.94 m Ux = Ucos e Us. using Acceleration ax = 0 ay . . 9 ¥1:0.199,9.18 speed of any time t v 1.04 ml. U* = vx Displacement of any time C.O. . W Sx % Ust , NOVEL Distance Haveled by the student living.g1.0 2.01.0.0890 11,40.2 No solution for t so she student H . using 1912 : usings using won't be able to catch up H. d. Vmn E 2 ad 9 H : Van 3.99 mls H 069.41 , e. . . . mm : 9 a 1 : 5.99mls R. (500) 0.179mls R 2999.64 D : Yma t D : 80 . 23 m CS Scanned with CamScanner

5. Given : v 300 mls 0. Given: Radius of circular track 0.20 solution: v, : solution: : a..." r (3006)n20))'(9.81)h h = (300 in 20) II/s. 2(9.81) total acceleration h,530.89 (5'+10.911) : 983 Y YY 105.45mlscc . 3.135 11/3' : : 300.99 mls 15 301 mls CS Scanned with CamScanner