#### To determine

**To find:** At what time the particle have a velocity of 20 m/s if the position of the particle is, s=t4−4t3−20t2+20t,t≥0.

#### Answer

The particle have the velocity of 20 m/s at *t* = 0 s or *t* = 5 s.

#### Explanation

The position function is, s=t4−4t3−20t2+20t,t≥0.

Find the derivative of s(t) and obtain the velocity.

v(t)=ddts(t)=ddt(t4−4t3−20t2+20t)=4t3−12t2−40t+20

Thus, the velocity is v(t)=4t3−12t2−40t+20,t≥0 (1)

Given that the velocity of particle is 20 m/s.

Substitute v(t)=20, and find the value of *t* as follows.

4t3−12t2−40t+20=204t3−12t2−40t=04t(t2−3t−10)=0t(t−5)(t+2)=0

Therefore, *t* = −2, 0 or 5.

Ignore the negative value as t≥0, the particle have the velocity of 20 m/s at *t* = 0 s or *t* = 5 s.

#### To determine

**To find:** At what time the acceleration of a particle become 0; significance of this time *t*.

#### Answer

The acceleration of a particle become 0 when t≈3.08 s, which shows that the acceleration changes from negative to positive and hence the velocity attains its minimum value.

#### Explanation

**Given:**

The position of the particle is, s=t4−4t3−20t2+20t,t≥0.

**Calculation:**

From part (a), v(t)=4t3−12t2−40t+20,t≥0.

Find the derivative of v(t) and obtain the acceleration.

a(t)=ddtv(t)=ddt(4t3−12t2−40t+20)=12t2−24t−40

Given that the acceleration of particle is 0 m/s.

Substitute a(t)=0, and find the value of *t* as follows.

12t2−24t−40=04(3t2−6t−10)=0

Use quadratic formula and simplify the terms further.

t=6±62−4(3)(−10)2(3)=6±36+1206=6±1566=1±1339

Therefore, the solutions are t≈3.08 since t≥0 ignore the negative value.

Thus, the acceleration of a particle become 0 when t≈3.08 s.

Notice that the acceleration changes from negative to positive and hence the velocity attains its minimum value.