Question

Two identical cars are traveling in long straight tracks placed side by side. A rope (which can be any length necessary) connects the centers of the two cars (ignore the dimensions of the cars). The positions of the cars change as r1= (2t^3+ 3t^2–10) m and r2= (5t^2–3t–10) m, where t is measured in seconds, for car 1 and car 2 respectively. Time begins being recorded when the rope is perpendicular to the track. The centers of the cars have a horizontal separation of 1 m at all

times.

I know that in order to find the velocity and acceleration for each I take the derivative twice (once for v, again for a); what I need help with is this:

1) Express in unit vector notation the relative position, velocity and acceleration of car 2 relative to car 1 at t = 3 s.

2) Determine the angle the rope makes relative to the orientation of the rope at t = 0 s and the length of the rope at t =3 s

Answer 1

Given

positions of the cars are

r1= (2t^3+ 3t^2–10) m and r2= (5t^2–3t–10) m,

the horizontal separtion is r = 1 m

relative position of the cars at t = 3 s is

r1(3s) = (2t^3+ 3t^2–10) m

= ((2*3^3+ 3*3^2–10) = 71 m

r2(3s) = (5*3^2-3*3-10) = 26 m

the relative postion is r2 - r1 = 71-26 = -45 m

the relative velocity at t= 3 s is  

V(1) = 6t^2+6t , V(2) = 10t -3

at t = 3 s is  

v(1) = 6*3^2 +6*3 = 54+18= 72 m/s, v(2) = 10*3 -3 = 27 m/s

V_rel = 72-27 = 45 m/s

now the acceleration is

a(1) = 12t +6 , a(2) = 10

at t= 3 s

a(1)= 12*3 + 6 = 42 m/s2 , a(2) = 10 m/s2

relative acceleration is a_rel = 42-10 = 32 m/s2

at time t = 3 s

the angle theta = arc sin ( 1/sqrt((-45)^2+1^2) = 1.27 degrees