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A particle P moves along the x-axis with its displacement at time t given by x = 6t^2 – t^3 + 1, where x is measured in metres and t in seconds. Find the velocity and acceleration of P at time t. Find the times at which P is at rest and find its position at these times.

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02:21

A particle P moves along the x-axis with its displacement at time t given by x = 6t^2 – t^3 + 1, where x is measured in meters and t in seconds. Find the velocity and acceleration of P at time t. Find the times at which P is at rest and find its position at these times.

02:26

A particle P moves along the x-axis with its displacement attime t given by x = 4t2 -t3 + 1, where x is measured in metres andt in seconds. Find the velocity and acceleration of P at time t.Find the times at which P is at rest and find its position at thesetimes.

02:36

#1:A particle starts at the origin (X; = 0) at time t = 0. Once it begins moving, the position of the particle as a function of time is given by:x(t) = (2.2 mls) t (-2.6 m/s2) t2 (-1.4 m/s?) t31. Find the particle’s velocity as a function of time, v(t) 2. Find the particle’s initial velocity; Vi, the velocity of the particle at time t = 0. 3. Find the particle’s acceleration as a function of time; alt). 4. Find the time that the particle reverses direction: (Hint: note that the function above only describes the motion for t > 0). 5. Find the location where the particle reverses direction. 6. Include a graph of the particles position, velocity, and acceleration as a function of time:

03:50

1. The position of a particle moving along a straight line isdefined by the relation x = t3 – 9t2 + 15t + 18, where x isexpressed in meters and t in seconds. Determine the time, position,and acceleration of the particle when its velocity becomeszero.2. The acceleration of a particle along a straight line is givenby the equation a = 4 – t2/9, where a is expressed in m/s2 and t inseconds. If the particle starts with zero velocity from a positionx = 0, find (a) its velocity after 6 s and (b) distance travelledin 6 s.

Transcript

Now, the given position in a first problem that is x of t is equal to 6 t square negative t to the power of 3 plus 1, where t is in second to find the velocity and acceleration a of t. The velocity v of t, rate of change of displacement or distance that is d o d t, 6 t square negative t cube plus 1 that is 12 times of t negative 3 times of t square. Now, the acceleration a of t is equal to d over d t of velocity v that is d over d t of 12 times of t negative 3 times of t square is equal to 12 negative 9 t. So, now, at rest velocity v of t has to be 0, then we have 12 times of t negative 3 times of t square has to be…

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