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if the position of a particle is s(t)=t^3-3t^2+4t-7, find the velocity and position of the particle when the acceleration is 0

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Answer:

When acceleration = 0,

velocity = 1

position = - 5

Step-by-step explanation:

The displacement of the particle is given by the equation

s(t) = t³ - 3t² + 4t - 7

The velocity equation can be determined by taking the first derivative which is the displacement per second

v(t) = s'(t) = d/dt( t³ - 3t² + 4t - 7) = 3t² - 6t + 4

The acceleration equation is obtained by taking the first derivative of v(t) which is the rate of change of velocity per second

a(t) = v'(t) = s"(t) = d/dt(3t² - 6t + 4) = 6t - 6

When acceleration is 0 we get a(t) = 0

6t - 6 = 0

6t = 6

t = 1 second

At t = 1, to find v(1) substitute t = 1 into

v(t) = 3t² - 6t + 4

v(1) = 3(1)² -6(1) + 4 = 3 - 6 + 4 = 1

s(1) = 1³ - 3 x 1² + 4 x 1 - 7

= 1 - 3 + 4 - 7

= 5 - 10

= -5

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