Given information:

The motion of the particle is given by the equation:

x=t2−(t−2)3

x=t2−(t3−(2)3−3t×2(t−2))x=t2−(t3−8−6t2+12t)x=t2−t3+8+6t2−12t

x=−t3+7t2−12t+8 __________ (1)

Where (x) in feet and (t) is in seconds.

We can obtain the velocity (v) at any time (t) by differentiating (x) with respect to (t),

Since, v=dxdt

v=ddt(−t3+7t2−12t+8)v= −3t2+14t−12 __________ (2)

The acceleration (a) can be obtained by differentiating again the above equation with respect to (t)

a=dvdta=ddt(−3t2+14t−12)

a=−6t+14 _________ (3)

When velocity is zero, v=0

From equation (2),

v  = −3t2+14t−12         or,0=  −3 t2+14t−12      or,−3 t2+14t−12=0t1=3.53st2=1.131s

Now, the position when t₁=3.53s.

x1=−t3+7t2−12t+8x1=−(3.53)3+7(3.53)2−12(3.53)+8x1=−43.98+87.22−42.36+8x1=95.22−86.34x1=8.88 ft

Now, the position when t₁=1.131s:

x2=−t3+7t2−12t+8x2=−(1.31)3+7(1.31)2−12(1.31)+8x2=−2.25+12−15.72+8x2=20−17.97x2=2.03 ft

Conclusion:

The two positions when velocity is zero x₁=8.88 ft and x₂=2.03ft.

Given information:

The motion of the particle is given by the equation:

x=t2−(t−2)3

x=t2−(t3−(2)3−3t×2(t−2))x=t2−(t3−8−6t2+12t)x=t2−t3+8+6t2−12t

x=−t3+7t2−12t+8 __________ (1)

Where (x) in feet and (t) is in seconds.

We can obtain the velocity (v) at any time (t) by differentiating (x) with respect to (t),

Since, v=dxdt

v=ddt(−t3+7t2−12t+8)v= −3t2+14t−12 __________ (2)

The acceleration (a) can be obtained by differentiating again the above equation with respect to (t):

a=dvdta=ddt(−3t2+14t−12)

a=−6t+14 _________ (3)

From equation (1), when t = 0s.

x1=−t3+7t2−12t+8x1=−(0)3+7(0)2−12(0)+8x1=8 ft

Again, from equation (1), when t=4s

x2=−t3+7t2−12t+8x2=−(4)3+7(4)2−12(4)+8x2=−64+112−48+8x2=120−112x2=8 ft

Conclusion:

The total distance travel by the particle is 8 ft.