Question

from O. 8 A particle of mass 2 kg moves in a straight line. At time t, its displacement from a fixed origin is x metres and its velocity is v ms-t. If the resultant force (in newtons) acting on the particle is: (a) 6 cost, and v=2 and x = 0 when t= 0, then find x in terms of t (b) 2 + 4x, and v= 2 when x = 0, then find v when x = 2 (c) 4-2v, and v=0 when t= 0, then find the time when v=1.

Answer 1

8. mass of particle = 2 kg At time to din pacement (a) 6 cost, and o=2 =x meters » du and x (0)=0 Force E= mass x acceleration » 6 cost = 2 X dv at = 3 cost dt Ult) = 3 X Sint + Ce (After integrating) Given: vrol=2 => 2 = so ult) = 3 Sint + 2 dxlt) = 3 Sint +2 뺥 Ci dt x(t) = 3 x.- Cost + 2t + C2 X(t)= - 3 Cost +2t +62 Given! x 10) = 0 => 0 3 -3 + C2 C2=3 so X(t) = -3cott2t+3 x(+1=-3 Cost + 2t+3 Ans; U (O) = 2, V () du d2x dta (b) Force = 2+ 4x Force mass x acci accn 244 x= 2 x d²x dt 2 + 4x = x = 2 x du dt 2 + 4x = 2 X ld di (:v=) > 2 + 4x= 2 ( ). It 22 = v do da vdu CBcanned with CamScanion tegrating (1+2x) dx = CamScanier teqrating both sides, we get

= (x + x2) + C1 where C, in constant. Now V (0) - 2 → LA uk 4 + O C,=-2 4 tc catas ) - 2 2 Atx=2 =16 v2= 2 n² + 2x + 4 V? - 8 +4+4 v=4 mis Ans: 14 ms1 (c) Force=4-20, ve (0=0, mass x aceh = 2 x dalt) = 4-2 dt dult) 2- Vit) 4-20 dt dult) dt - V 2-V ci 2 et integrating both sides, we get -In (2-1) = t + CA where C, in constant Vlo)= 0 => - ln2 so -ln (2-0) t-lna In a- ln (2-1)=t ln ( 22 )=t 2-1 V=2(1-é-t) Now Val en2 = 0.69 seconds. Ami In 2 or 0.69 seconds (žu) 1 2 (1-e-t) steln - > t CSScanned with damScanner