The position function of a particle moving along the x-axis is given by x(t)=t^3-12t^2-36t+20 0≤t≤8

Question

(a) Find the velocity and acceleration of the particle.

(b) Find the open t-intervals when the particle is moving to the left.

(c) Find the velocity of the particle when the acceleration is 0.

(d) Find when the particle is speeding up.

(e) Find when the particle is furthest to the right.

Very confused on part c and e can someone please explain this to me?

Answer 1

(a) v(t) = x'(t) = 3*t²-24*t-36 a(t)=v'(t)=x''(t) = 6*t-24

(b) The particle is moving to the left when the velocity is less than 0.

e.g. v(t)<0 3*t²-24*t-36<0

t²-8*t-12<0

t=4+2√7 only, since 4-2√7 is not in the specified interval

The interval (0,4+2√7) gives a negative velocity.

(c) a(t)=0=6*t-24

t=4 ; The acceleration is 0 when time t=4.

The velocity at time t=4 is then:

v(4)=3*4²-24*4-36 = 48-96-36 = -84

(d) The particle is speeding up when the acceleration is greater than 0.

a(t)=6*t-24>0

t > 4

The interval where it is speeding up is (4,8)

(e) The particle is farthest to the right when the distance is at a maximum. This can be found when the derivative of the distance (velocity) is zero, or at either ends of the interval.

v(t)=3*t²-24*t-36 = 0

t=4+2√7

Checking the distance at the beginning of the interval, end of the interval and t=4+2√7, we get:

x(0)=20

x(8)=-524

x(4+2√7)=-548.32.

Therefore, the distance is at a maximum at time t=0.