Question

PE 1.pdf The motion of a jet plane just after landing on a runway is described by the a-t graph. Determine the time t' when the jet plane stops. Construct the v-t and s-t graphs for the motion. Here s=0, and v= 150 m/s when t=0. DODDDDDDD a (m/s²) 10 20 ť 1 (S) --10 -20

Answer 1

Solution, a(m/32) Basic Formula ds - ds=vat dt (0,0) 10 20 t' du - a gdy=adt t(s) de- -Ib -20 Given That when too s=0 and 1= 150m/see In the range of Ox£< 10 acceleration so in do= adt initially at t-o; w=150 say at timet; velocity is el 9 hence; (do. Just is sore-sole 150 v-150=0 = 150 m/see so in a range of Ostslo; v=150 honce at t=10; 1=150 m/see, Next in between lost<2o we can see the Curve nature is Let us forom the strenght line equertion we Com see at t=10 ; a=-20 a Straight line als at t= 20; a = -10

a- 30 t-10 (-20) -20-(-10) 10-20 dd20 t-10 -10 -10 09, a+ 20=t-10 ore, a=t-10-20 or, fore the la= t-30 acceleration function tange of 1024-20 Now; initially at t-to; v= 150m (see say at time t", velocity is "r" In this teange t fare adt (de= ((e-30)dt 150 10 150 2 t [w] 150 = [+ - 2 3 10 13043 סן v-150= t² 102 - [30t- 300 2 2 U-150 = - 50- 30t +300 2 v= t2 -30t +400 2 so in reange of 105 t=20; = €²-30t +400 hence at t= 20; e=0 m/sec

Next, In a range of 20 t<t; the acceleration is constant as a--10 Initially at t= 20; 2=0 m/see; sand at timet" velocity is "ne" hence t O 20 Sae - Sadt Sav=ficat 20 t - 0= [-104] 20 = -lot +200 at t=t the jet plane stops. hence, U=0 at tt in range of zostt 30 O= -100 4200 10t'=200 t = 20 see. so at 20 sec. the jet plane will stop. rent diagram based on this range wise calculation, re(m/sec) 1=150 150 th _ 30t+400 → tlsee) t= 20

o<t10; when timet: distance is 5 (sara) 5 t Now fore Sds=fredt Sisodt [s-o] = [1sot t oras and in this range O=150 O 3= 150t 30 at; t=10; 3= 1500m. Now for 10< t <20 When tine "t" distance is "s" (sub) In this thonge = €² 2 .-30t thoo s odt 3 t so ds = 10 1500 8 ds = SE2_30t thooldt ID 1500 t [579,500 Choot 2 10 103 6 2 15tt 1500 + 4oot-4000 5-150o = t3 6 sa t3 6 2 15t +yoot 3500 3 thired trenge at so, t=20; 5= 2166. 667m. Now we do not need the 20 2txt since we already found out that the jet at t t=20 will stop

hence no thired trange will exist. Therefore set diagram based on teange- wise calculation 3(m) 21666667 2 1500 - 3500 15th thoot 3 3=150t A(sec) t-to t=20