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We now know that taking the derivative of a function will give us the slope, or the instantaneous rate of change of the function. So what if we take the derivative of a function that models the position of some object moving along a line? It gives us its velocity! And if we differentiate its velocity function? It gives us its acceleration! In this section, we will study the relationship between position, velocity and acceleration using our knowledge of differential calculus.

motion along a straight line

$s(t)$: position

$v(t)=s'(t)$: instantaneous velocity

$a(t)=v' (t)=s''(t)$: acceleration

$s(t)$: position

$v(t)=s'(t)$: instantaneous velocity

$a(t)=v' (t)=s''(t)$: acceleration

- 1.The position of a particle moving along a straight line is given by:

$s(t)=t^3-12t^2+36t-14$

where t is measured in seconds and s in meters.a)Find the position of the particle at: t=0,1,2,3,4,5,6,7, and draw a diagram to illustrate the positions of the particle.b)Find the velocity at time t.c)Find the velocity of the particle at: t=1,3,5,7.d)When is the particle:

i) at rest?

ii) moving forward (that is, moving in the positive direction)?

iii) moving backward (that is, moving in the negative direction)?

Draw a diagram to illustrate the motion of the particle.e)Find the total distance traveled by the particle during the first 7 seconds.f)Find the acceleration at time t.g)Compare the velocity and acceleration of the particle at: t=1,3,5,7, and determine whether the particle is speeding up or slowing down at each instant.h)Graph the position, velocity, and acceleration functions for

$0 \leq t \leq 7.$i)When is the particle:

i) speeding up?

ii) slowing down?

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