Question

5. (15 pts] A force acting on an object is proportional to the square root of the distance the object moves. The equation for the distance r is: 19 = kyr, for some constant k. The object starts at t=0 with position r = 0 and velocity v = 0. The object moves four meters in the first second. That is, r(t = 1) = 4. (a) Find the object's velocity as a function of position, v(r). (b) Find the object's position as a function of time, r(t). (c) Find its acceleration at time t = 2 seconds.

Answer 1

Comment in case of any doubt.

- linen 2 = Kh ) vodu iz Kir - » var = kve dr Jude= Skilęde (IDŹ liven at t=0, r=0,v=o, so from -6) above for r=0 and rzo, we get o= o+c =) C=0 So, we have v = 2 ks =) V² 4 Kisha - y = ( x ) = 2 Answer by a c = 24 de 3 - 455834 (K is found in part (C) below) de -21 dt > Se de fant dat n re - At t=0, r=0, which gives c=0 1. Thus, 2.4 t = aut r = 11 ) " ) 2 = 1 + 1 (c) At t= l r =4, so 4 1 13" = 42 = 4x 4 4 2 3 K = 21 1. So, we get r=44 ? Now, at t= 2,2=4 (2) = 64 So, as dr. alte = kJr = (24) J64 = 192 m/s2 16 5 444