Question

The position of a particle moving along the x axis is given in
meters by x = 3.0t^{2} – 1.0t^{3}, where t is in
seconds. (a.) At what time does the particle reach its maximum
positive x position? (b.) What total length of path does the
particle cover in the first 4.0 sec? (c.) What is its displacement
during the first 4.0 sec? (d.) What is the particle’s speed at the
end of the first 4 sec? (e.) What is the particle’s acceleration at
the end of the first 4 sec?

Answer #1

Velocity is defined as the derivative of position and similarly, acceleration is defined as the derivative of velocity.

At maximum positive x position, particle velocity becomes zero and it changes ots direction.

Displacement is the difference between final and initial position.

The position of a particle moving along the x axis
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ct2 − bt3, where x
is in meters and t in seconds.
For the following, let the numerical values of c and
b be 5.1 and 1.5, respectively. (For vector quantities,
indicate direction with the sign of your answer.)
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please do 1,2 and 3 thanks
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