Chapter 3.3, Problem 41PP

(a)

To determine

The velocity of the brick after 4.0s.

Answer to Problem 41PP

  $39.2\text{m/s}$

Explanation of Solution

Given:

A brick is dropped from a high scaffold.

Formula used:

Final velocity with constant acceleration is given by,

  $v_{final}=v_{initial}+a\Delta t$

Where, $\Delta t$ is the time taken, a is acceleration and v represents velocity.

Calculation:

Since, the brick is dropped from scaffold, thus its initial velocity will be 0, that is:

  $v_{initial}=0\text{m/s}$

Acceleration acting on it will be due to gravity only. Thus,

  $a=9.8{\text{m/s}}^{\text{2}}$

Then, using the above mentioned formula for the final velocity,

  $\begin{array}{c}{v}_{final}=v_{initial}+a\Delta t\\ =0+\left(9.8\times 4.0\right)\\ =39.2\text{m/s}\end{array}$

Conclusion:

Thus, the final velocity of the brick after 4.0s is $39.2\text{m/s}$ .

(b)

To determine

The distance travelled by the brick in 4.0s.

Answer to Problem 41PP

  $78.4\text{m}$

Explanation of Solution

Given:

A brick is dropped from a high scaffold.

Formula used:

The distance traveled by an object with a constant acceleration is given by

  $d_{final}=d_{initial}+v_{initial}\Delta t+\frac{1}{2}a{\left(\Delta t\right)}^2$

Here, d represents displacement, v represents velocity, t represents time and a is the acceleration.

Calculation:

Since, the brick is dropped from scaffold, thus its initial velocity will be 0, that is,

  $v_{initial}=0\text{m/s}$

And, only acceleration acting on it will be due to gravity only. Thus,

  $a=9.8{\text{m/s}}^{\text{2}}$

Then, using the above-mentioned formula for the distance covered by the brick in 4.0s,

  $\begin{array}{c}{d}_{final}=d_{initial}+v_{initial}\Delta t+\frac{1}{2}a{\left(\Delta t\right)}^2\\ =0+0\left(4\right)+\frac{1}{2}\left(9.8\right){\left(4.0\right)}^2\\ =78.4\text{m}\end{array}$

Conclusion:

Thus, the distance covered by the brick in 4.0s is $78.4\text{m}$ .