Soshi K.

asked • 10/15/20# 1-D Kinematics of Particles

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/𝑎^{2} 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.

## 1 Expert Answer

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})t^{2}, for B V_{B}(t)= 0.25 +0.5t; then for B after integration S_{B}(t)=0.25t+0.25t^{2}. 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.

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Petros H.

10/21/20