1-D Kinematics of Particles

Question

Particle 𝐴, which is initially at rest, wants to meet Particle 𝐵, however they are 𝑋𝑌𝑍 ft away from one another. The motion of particle 𝐴 heading toward 𝐵 is defined by the relation: 𝑠 = 1/𝑎² where 𝑠 and 𝑎 are expressed in ft and ft s 2 , respectively. On the other hand, Particle 𝐵 travels toward 𝐴 with an initial velocity of 0.25 ft s that increases at the rate of 0.50 ft s 2 . With the motion of each particle described above, determine when and where they will meet each other in reference to Particle 𝐴’s initial position.

Answer 1

Please, try to present the problem more understandable. For example, is XYZ a 3D distance, or it is just your symbol? Next, the relation S=a^-2 , maybe is S(t)=(a^-2)t^2 , since without time the formula cannot describe a motion.

If so, the both motions are accelerated: for A S_a(t)=(a^-2)t², for B V_B(t)= 0.25 +0.5t; then for B after integration S_B(t)=0.25t+0.25t². Now, you will have three unknowns (t, S_A , and S_B ) and three equations (S_A, S_B, and (S_A+S_B)=XYZ). By keeping in mind that t_A=t_B at the meeting moment you can easy solve the problem.