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A particle is moving along the x-axis. Its position as a function of time is given as x=bt-ct^2a) What must be the units of the constants b and c, if x is in meters and t in seconds?b) At time zero, the particle is at the origin. At what later time t does it pass the origin again?c) Derive an expression for the x-component of velocity.d) At what time t is the particle momentarily at rest?e) Derive an expression for the x-component of the particles acceleration, ax

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hopemichael95

Answer:

We are given x= bt +ct²

So

A. bxt= m

Because m/s*s= m

So b= m/s and c= m/s²

B.

x= bt-ct²

So at x=0 t=0

x=0 t= 2

We have

bt = ct² so t = b/c at x= 0

So b-2ct= 0

B. To find velocity we use

dx / dt = b - 2 Ct

C. At rest wen V= 0

We have t= b/2c

D. To find acceleration we use

dv / dt = - 2C

When We are given x= bt +ct²

Then A. bx.t = m

  • Just Because of the  m/s*s= m
  • So that,  b= m/s and c= m/s²
  • Now B. x= bt-ct²
  • the So at x=0 t=0
  • After that x=0 t= 2
  • Now We have bt = ct² so t = b/c at x= 0
  • After that So b-2ct= 0

Then B. To find velocity we use that is

  • Then dx / dt = b - 2 Ct

Then C. At rest wen V= 0

  • We have t= b/2c

Now D. To find acceleration we use

  • Now the answer is dv / dt = - 2C

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