Dr. Z’s Math251 Handout #13.4 [Motion in Space: Velocity and Acceleration]By Doron ZeilbergerProblem Type 13.4a: Find the velocity, acceleration, and speed of a particle with the givenposition function.r(t) = x(t) i + y(t) j + z(t) k .Example Problem 13.4a: Find the velocity, acceleration, and speed of a particle with the givenposition function.r(t) = t2i + ln t j + t k .Steps Example1. The velocity is r0(t) and the accelera-tion is r00(t).1. v(t) = r0(t) = (t2)0i + (ln t)0j + t0k= 2t i + 1/t j + k.a(t) = r00(t) = (2t)0i + (1/t)0j + 10k =2 i − (1/t2) j .2. To find the speed, take the magnitudeof the velocity, i.e. compute |v(t)|2. the speed is |v(t)| = |2t i + 1/t j + k| =p(2t)2+ (1/t)2+ 12=p4t2+ 1/t2+ 1.Problem Type 13.4b: A force with magnitute F N acts on a body of mass m in the directonhd1, d2, d3i. The object starts at the (x0, y0, z0) with initial velocity v(0) = hv1, v2, v3i. Find itsposition function and its speed at time t.Example Problem 13.4b: A force with magnitute 300N acts on a body of mass 100 kg in thedirecton h1, 2, 2i. The object starts at the (1, 2, 3) with initial velocity v(0) = h0, 1, −1i. Find itsposition function and its speed at time t.Steps Example1. Find the unit direction vector bydividing hd1, d2, d3i by its length. To getthe force vector, multiply this vector bythe magnitude of the force F .1. |h1, 2, 2i| =√12+ 22+ 22= 3. Sothe direction of the force is (1/3)h1, 2, 2i=h1/3, 2/3, 2/3i. The force is F = 300h1/3, 2/3, 2/3i,so F = h100, 200, 200i.12. Set-up Newton’s Second LawF = mr00(t) .2.h100, 200, 200i = 100r00(t) .sor00(t) = h1, 2, 2i .3. Integrate with respect to t to get v(t) =r0(t), and don’t forget the arbitrary con-stant vector. Then plug-in t = 0 to getit.3.v(t) = r0(t) =Zh1, 2, 2idt = ht, 2t, 2ti+C .But v(0) = h0, 1, −1i, so C = h0, 1, −1iand we getv(t) = r0(t) = ht, 2t, 2ti+h0, 1, −1i = ht, 2t+1, 2t−1i .4. To get the position vector r(t) inte-grate the velocity vector v(t) that youfound in step 2, once again not forgettingthe arbitrary constant vector.4.r(t) =Zht, 2t+1, 2t−1idt = ht2/2, t2+t, t2−ti+CWhen t = 0, r(0) = h1, 2, 3i, so C =h1, 2, 3i, andr(t) = ht2/2, t2+t, t2−ti+h1, 2, 3i = ht2/2+1, t2+t+2, t2−t+3i .This is the position function.5. To get the speed, compute |v(t)|, anexpression in t.5.|v(t)| = |ht, 2t + 1, 2t − 1i|=pt2+ (2t + 1)2+ (2t − 1)2=p9t2+ 2