**\[ v(t) = 3t -8, 0 \leq t \leq 3 \]**

**(a) Find the displacement. **

**(b) Find the distance traveled by the particle during the given time interval.**

The aim of the **question** is to understand how to **calculate** the **displacement** and the **distance** covered by the **moving** particle in the given **velocity** and the **time** interval.

**Displacement** is the change in the **position** of an object. Displacement is a **vector** and has **direction** and **magnitude.** It is denoted by the **arrow** that goes from the starting **position** to the **final.**

The total **distance** traveled is **calculated** by finding the **area** under the **velocity** curve from the given **time** interval.

## Expert Answer

**Part a**

Since $v(t) = x'(t)$ where x(t) is the **displacement** function, then the **displacement** over the interval $[a,b]$ given $v(t)$ is $\int_a^b v(t) dt$, It is given that $v(t)= 3t-8$ and the **interval** is $[0,3]$, so the **displacement** is:

\[= \int_0^3 v(t) dt \]

\[= \int_0^3 (3t-8) dt \]

Applying the **integration:**

\[= \left( \dfrac{3} {2} t^2 – 8t \right)_0^3 \]

Inserting the **limits:**

\[= \left( \dfrac{3} {2} (3)^2 – 8(3) \right) – \left( \dfrac{3} {2} (0)^2 – 8(0) \right) \]

\[= \dfrac{3} {2} (9) – 24 \]

\[= \dfrac{27} {2} – 24 \]

\[= -10.5\]

**Part b**

Total **distance** traveled = $\int_a^b |v(t)| dt$ for an **interval** $[a,b]$. You then determine where $v(t)$ is **positive** and **negative** so you can rewrite the **integral** to have absolute **values.**

Setting $v(t) = 0$ and **solving** for $t$ gives:

\[ 0= 3t-8 \]

\[8= 3t \]

\[t= \dfrac{8} {3} \]

Since $t=1$ lies in the **interval** $[0, \dfrac{8}{3}]$ and $v(t) = 3(1)-8$.

That is $-5$ and $< 0$, then $v(t)<0$ for $[0, \dfrac{8}{3}]$.

Since $t=2.7$ lies in the **interval** $[\dfrac{8}{3}, 3]$ and $v(t) = 3(2.7)-8$.

That is $0.1$ and $> 0$, then $v(t)>0$ for $[\dfrac{8}{3}, 3]$.

To break **apart** the absolute **value,** you then need to **write** the integral as a sum of **integrals** over each integral where the **interval** with $v(t)<0$ has a negative in **front** and the interval with $v(t)>0$ has a **plus** front:

\[ \int_0^3 |v(t)| dt = \int_0^3 |3(t)-8| dt \]

\[ – \int_0^{\dfrac{8} {3}} (3(t)-8) dt + \int_{ \dfrac{8} {3}}^3 (3(t)-8) dt \]

\[ – \left( \dfrac{3}{2} t^2 – 8t \right)_0^{\dfrac{8} {3}} + \left( \dfrac{3}{2} t^2 – 8t \right)_{\dfrac{8} {3}}^3 \]

\[ – \left[ \left( \dfrac{3}{2} (\dfrac{8} {3})^2 – 8(\dfrac{8}{3}) \right) – \left( \dfrac{3} {2} (0)^2 – 8(0) \right) \right] + \left[ \left( \dfrac{3}{2} (3)^2 – 8(3) \right) – \left( \dfrac{3} {2} (\dfrac{8}{3})^2 – 8(\dfrac{8} {3}) \right) \right] \]

By solving the **above** expression:

\[= \dfrac{32}{3} – \dfrac{21}{2} + \dfrac{32} {3} \]

\[= \dfrac{65} {6} \]

\[= 10.833\]

## Numerical Answer

**Part a: Displacement** = $-10.5$

**Part b: **Distance **traveled** by the particle is = $10.833$

## Example

Find the **displacement** if the velocity is given as:

\[ v(t)= 6- t, 0 \leq t \leq 6 \]

\[= \int_0^6 v(t) dt \]

\[= \int_0^6 (6-t) dt \]

Applying the **integration:**

\[= (6t – \dfrac{1}{2}t^2 )_0^6 \]

Inserting the **limits:**

\[= (6(6) – \dfrac{1}{2} (6)^2) – ((0)t – \dfrac{1}{2} (0)^2 ) \]

\[= (36 – 18) \]

\[= 18 \]