#### To determine

**a)**

**To find:**

The displacement.

#### Answer

-32 m

#### Explanation

**1) Concept:**

vt=s't ;vt& st are the velocity and the displacement function of the particle respectively.

**2) Given:**

vt=3t-5, 0≤t≤3

**3) Calculation:**

The velocity of the particle moving along a line,

vt=3t-5, 0≤t≤3

To find the displacement of particle st, use vt=s't, i.e.,

vt=s't=3t-5, 0≤t≤3

So by using the Net change theorem s(x)-s(0)=∫0xs'(t)dt=∫0x(3t-5)dt Integrating and rearranging, it becomes

sx=3t22-5t 0x+s(0)

=3x22-5x -0 = 3x22-5x Thus the displacement at time t is given by s(t) = 3t22-5t

So displacement at time 3 seconds is given by s3= =3·322-5·3

By simplifying,

st=-32 Therefore, the particle is 32 meters to the left.

**Conclusion:**

Therefore, the displacement is -32 m

#### To determine

**b)**

**To find:**

The distance traveled by the particle during the given time interval.

#### Answer

416m

#### Explanation

**1) Concept:**

i) Total distance travelled: ∫t2t1⃓v(t)⃓dt ; t1≤t≤t2

ii) Net Change theorem: The integral of a rate of change is the net change.

∫abF'xdx=Fb-Fa

**2) Given:**

vt=3t-5, 0≤t≤3

**3) Calculation**

The velocity of the particle moving along a line,

vt=3t-5, 0≤t≤3

Speed is the rate of change of distance travelled. Speed is given by the magnitude of velocity. That is |v(t)|. So to find total distance travelled we shall integrate speed within the given interval. Thus the distanced travelled in first three second is

∫03⃓v(t)⃓dt=∫03⃓3t-5⃓dt

In the interval 53,3 we have vt≥0 (the particle moves to the right) In the interval 0,53 we have v(t)≤0(the particle moves to the left).

Since the velocity is negative on 0,53 , write the indefinite integral ∫03⃓3t-5⃓dt as

∫03⃓3t-5⃓dt=∫053-(3t-5)dt+∫533(3t-5)dt

By simplifying,

=∫053(-3t+5)dt+∫533(3t-5)dt

By integrating,

=-3t22+5t053+3t22-5t533

By using the fundamental theorem of calculus,

=-3·(53)22+5·53-0+3·322-5·3-3·(53)22-5·53

By simplifying,

=-256+253-32-256-253

=-256+253-32-256+253

=503-506-32

=1006-506-32

=506-32

=416

Therefore, the total distance travels is 416 meters.

**Conclusion:**

Therefore, the total distance travels is 416 meters.