Chapter 1.1, Problem 3E

To determine

To find: the velocity of the object after 3 seconds of falling with a constant acceleration.

Answer to Problem 3E

  $V_{\text{final}}=96\text{ft}/\text{s}$

Explanation of Solution

Given information:

The given equation of falling the object is $y=16t^2$ after $t$ seconds of falling.

The average velocity of the object after 3 seconds of falling is $48\text{ft}/{\text{s}}^2$ .

Formula used:

Average velocity after $t$ second of a moving object with constant acceleration can be found by the following formula:

  $V_{avg}=\frac{\text{Final velocity}+\text{Initial velocity}}{\text{2}}$

Calculation:

Calculate the initial position of the object by substituting 0 for $t$ in the equation of motion.

  $\begin{array}{c}y=16{\left(0\right)}^2\\ =0\end{array}$

Now, calculate the velocity of the object after 3 seconds of falling as shown:

  $\begin{array}{c}{V}_{avg}=\frac{\text{Final velocity}+\text{Initial velocity}}{\text{2}}\\ 48=\frac{\text{Final velocity}+\text{0}}{2}\\ \text{Final velocity}=48\times 2\\ =96\text{ft}/\text{s}\end{array}$

Hence, the average velocity of the object during first 3 seconds of fall is $96\text{ft}/\text{s}$ .