Two particles movealong an x axis. The position of particle 1 is given by x = 6.

Question

Two particles movealong an x axis. The position of particle 1 is given by x= 6.80t2 +3.00t + 2.00 (in meters and seconds); the acceleration ofparticle 2 is given by a = -8.00t (in meters perseconds squared and seconds) and, at t = 0, its velocityis 19 m/s.When the velocities of theparticles match, what is their velocity?

1 m/s Two particles movealong an x axis. The position of particle 1 is given by x= 6.80t2 +3.00t + 2.00 (in meters and seconds); the acceleration ofparticle 2 is given by a = -8.00t (in meters perseconds squared and seconds) and, at t = 0, its velocityis 19 m/s.When the velocities of theparticles match, what is their velocity?

Two particles movealong an x axis. The position of particle 1 is given by x= 6.80t2 +3.00t + 2.00 (in meters and seconds); the acceleration ofparticle 2 is given by a = -8.00t (in meters perseconds squared and seconds) and, at t = 0, its velocityis 19 m/s.When the velocities of theparticles match, what is their velocity?

Explanation / Answer

This is almost more of a calculus question then a physicsquestion. Basically, what you are going to want to do is take thederivative of the position function to find the velocity of thefirst particle. Using the power rule (if f(x)=ax^n , thenf'(x)=nax^n-1) I got a derivative of 13.6t+3 so that is thevelocity of the 1st particle For the second particle you need an anti-derivative. if youhave ax^n in this context, to find the anti-derivative it is(ax^(n+1))/(n+1) I got -4t^2+C (the C stands for a constant,because you do not know if there was a constant in the originalequation) since they give you that v(0)=19, you know that-4(0)^2+C=19, so C=19, giving you a final equation of-4t^2+19. Now all you need to do is set 13.6t+3 equal to -4t^2+19, andthat will give you an answer. Hope that helped

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.