Chapter 3.3, Problem 42PP

(a)

To determine

The velocity of the brick after 4 s.

Answer to Problem 42PP

The velocity of the brick after 4 s is $39.2\text{m}/\text{s}$ .

Explanation of Solution

Given Info:

The initial velocity of the brick is, $u=0\text{m}/\text{s}$

The time to find the velocity of brick is, $t=4\text{s}$

Formula Used:

The expression to calculate the velocity of the brick is,

$v=u+gt$

Here,

• $v$is the velocity of the brick.
• $g$is the gravitational acceleration.

Calculation:

Substitute all the values in the above expression.

$\begin{array}{c}v=0\text{m}/\text{s}+\left(9.8\text{m}/{\text{s}}^2\right)\left(4\text{s}\right)\\ =39.2\text{m}/\text{s}\end{array}$

Thus, the velocity of the brick after 4 s is $39.2\text{m}/\text{s}$ .

(b)

To determine

The distance covered by the brick in 4 s.

Answer to Problem 42PP

The distance covered by the brick in 4 s is $78.4\text{m}$ .

Explanation of Solution

Formula Used:

The expression to calculate the distance covered by the brick is,

$h=ut+\frac{1}{2}g{t}^2$

Here,

• $h$ is the distance covered by the brick.
• $u$is the initial velocity of the brick.

Calculation:

Substitute all the values in the above expression.

$\begin{array}{c}h=\left(0\text{m}/\text{s}\right)\left(4\text{s}\right)+\frac{1}{2}\left(9.8\text{m}/{\text{s}}^2\right){\left(4\text{s}\right)}^2\\ =78.4\text{m}\end{array}$

Thus, the distance covered by the brick in 4 s is $78.4\text{m}$ .