# Question 75ba1

Jun 28, 2015

Here's how you'd go about solving this one.

#### Explanation:

You know that your object starts from rest, which means that tis initial velocity will be equal to zero.

Moreover, it has an uniform acceleration of $4 {\text{m/s}}^{2}$. To determine its displacement, you can use this equation

$d = {v}_{0} \cdot t + \frac{1}{2} \cdot a \cdot {t}^{2}$, where

$t$ - the object's time of travel;
$a$ - its acceleration.

$d = {\underbrace{{v}_{0}}}_{\textcolor{b l u e}{\text{=0}}} \cdot t + \frac{1}{2} \cdot a \cdot {t}^{2} = \frac{1}{2} \cdot a \cdot {t}^{2}$

Point (1)

After 5 seconds, the object's displacement will be

d = 1/2 * 4"m"/cancel("s"^2) * 5^2cancel("s"^2) = color(green)("50 m")#

Point (2)

To determine how much the object travelled in the ${5}^{\text{th}}$ second of movement, you need to subtract the distance it travelled after 5 seconds from the distance it travelled after 6 seconds.

$\Delta {d}_{5 , 6} = {d}_{6} - {d}_{5}$

${d}_{6} = \frac{1}{2} \cdot 4 \text{m"/cancel("s"^2) * 6^2cancel("s"^2) = "72 m}$

${d}_{5} = \text{50 m}$

$\Delta {d}_{5 , 6} = 72 - 50 = \textcolor{g r e e n}{\text{22 m}}$

Point (3)

The same method can be used to determine the distance the object travelled in its ${8}^{\text{th}}$ second of movement.

$\Delta {d}_{8 , 9} = {d}_{9} - {d}_{8}$

${d}_{9} = \frac{1}{2} \cdot 4 \text{m"/cancel("s"^2) * 9^2cancel("s"^2) = "162 m}$

${d}_{8} = \frac{1}{2} \cdot 4 \text{m"/cancel("s"^2) * 8^2cancel("s"^2) = "128 m}$

Therefore,

$\Delta {d}_{8 , 9} = 162 - 128 = \textcolor{g r e e n}{\text{34 m}}$