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mash [69]
3 years ago
13

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
Physics
1 answer:
lisabon 2012 [21]3 years ago
3 0

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

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Answer:

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Explanation:

Velocity is the change in displacement over change in time, this makes it a rate of change. It can be positive or negative because it is a vector quantity. It uses the units m/s because that is a displacement unit over a time unit.

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Answer:

f = 3.1 kHz

Explanation:

given,

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