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The function ​s(t)=−t3+3t+3 gives the distance from a starting point at time t of a particle moving along a line. Find the velocity and acceleration functions. Then find the velocity and acceleration at t=0 and t=1. Assume that time is measured in seconds and distance is measured in centimeters. Velocity will be in centimeters per second​ (cm/sec) and acceleration in centimeters per second per second ​(cm/sec2​).

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Answer:

v(t)=3t²+3, and a(t)=6t

v(0)=3cm/sec, and a(0)=0cm/sec²

v(1)=6cm/sec, and a(1)=6cm/sec²

Explanation:

Find velocity and acceleration functions ( v(t) and a(t) )

(Relevant background: Let's say f(t) is a function of y with respect to x. Think of the derivative of this function, f'(t) , as representing the rate at which y changes with respect to x)

s(t) is a function of distance with respect to time. Therefore, s'(t) - the derivative of this - represents the rate at which distance changes with time, which is just the definition of velocity. So we can say velocity v(t) = s'(t).

s(t)=t³+3t+3

s'(t)=v(t)=3t²+3

Similarly, if v(t) is a function of speed with respect to time, then v'(t) represents the rate at which speed changes with time, which is acceleration. So we can say that acceleration a(t)=v'(t)=s''(t)

v(t)=3t²+3

v'(t)=a(t)=6t

Find velocity and acceleration at t=0 and t=1

t=0

v(t)=3t²+3

v(0)=3(0²)+3

v(0)=3 cm/sec

a(t)=6t

a(0)=6(0)

a(0)=0 cm/sec²

t=1

v(t)=3t²+3

v(1)=3(1²)+3

v(1)=3+3

v(t)=6 cm/sec

a(t)=6t

a(1)=6(1)

a(1)=6 cm/sec²

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