A football quarterback runs 15.0 m straight down the playing field in 2.50 s. He is then hit and pushed 3.00 m straight backward in 1.75 s. He breaks the tackle and runs straight forward another 21.0 m in 5.20 s. Calculate his average velocity
(a) for the first time interval,
(b) for the second time interval,
(c) for the third time interval, and
(d) for the entire motion.
The Correct Answer and Explanation is :
To calculate the average velocity, we use the formula:
[
v_{avg} = \frac{\Delta x}{\Delta t}
]
where:
- (\Delta x) is the displacement,
- (\Delta t) is the time interval.
We will use this formula for each of the three time intervals and for the entire motion.
(a) Average velocity for the first time interval
In the first interval, the quarterback runs forward 15.0 meters in 2.50 seconds. Since the motion is straight down the field, the displacement is 15.0 m in the forward direction.
[
v_{avg,1} = \frac{\Delta x_1}{\Delta t_1} = \frac{15.0 \, \text{m}}{2.50 \, \text{s}} = 6.00 \, \text{m/s}
]
(b) Average velocity for the second time interval
In the second interval, the quarterback is pushed backward by 3.00 meters in 1.75 seconds. This motion is in the opposite direction of his initial run, so the displacement is negative.
[
v_{avg,2} = \frac{\Delta x_2}{\Delta t_2} = \frac{-3.00 \, \text{m}}{1.75 \, \text{s}} = -1.71 \, \text{m/s}
]
(c) Average velocity for the third time interval
In the third interval, the quarterback runs forward 21.0 meters in 5.20 seconds. Since he is running forward, the displacement is positive.
[
v_{avg,3} = \frac{\Delta x_3}{\Delta t_3} = \frac{21.0 \, \text{m}}{5.20 \, \text{s}} = 4.04 \, \text{m/s}
]
(d) Average velocity for the entire motion
For the entire motion, we add up the total displacement and total time. The total displacement is:
[
\Delta x_{\text{total}} = 15.0 \, \text{m} – 3.00 \, \text{m} + 21.0 \, \text{m} = 33.0 \, \text{m}
]
The total time is:
[
\Delta t_{\text{total}} = 2.50 \, \text{s} + 1.75 \, \text{s} + 5.20 \, \text{s} = 9.45 \, \text{s}
]
Now, we can calculate the average velocity for the entire motion:
[
v_{avg,\text{total}} = \frac{\Delta x_{\text{total}}}{\Delta t_{\text{total}}} = \frac{33.0 \, \text{m}}{9.45 \, \text{s}} = 3.49 \, \text{m/s}
]
Final Answers:
- (a) (v_{avg,1} = 6.00 \, \text{m/s})
- (b) (v_{avg,2} = -1.71 \, \text{m/s})
- (c) (v_{avg,3} = 4.04 \, \text{m/s})
- (d) (v_{avg,\text{total}} = 3.49 \, \text{m/s})
Explanation:
In this problem, we are dealing with multiple intervals of motion in different directions. For each interval, we calculate the average velocity by dividing the displacement by the time taken. The displacement for the second interval is negative because the quarterback is pushed backward, which affects the average velocity for that part of the motion. The average velocity for the entire motion considers the total displacement (forward and backward) over the total time. The key idea is that velocity is a vector quantity, meaning direction matters, and the average velocity accounts for both the magnitude and direction of the motion.