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Answer :
The magnitude of the average acceleration is [tex]\( 1.41 \, \text{m/s}^2 \).[/tex]
1. Convert speed to meters per second:
Given speed: [tex]\( 157 \, \text{mi/h} \)[/tex]
Conversion factors: [tex]\( 1 \, \text{mi} = 1609.34 \, \text{m} \) and \( 1 \, \text{h} = 3600 \, \text{s} \)[/tex]
[tex]\[ 157 \, \text{mi/h} \times \frac{1609.34 \, \text{m}}{1 \, \text{mi}} \times \frac{1 \, \text{h}}{3600 \, \text{s}} = 70.2 \, \text{m/s} \][/tex]
2.Calculate average acceleration:
Initial speed [tex]\( u = 0 \, \text{m/s} \)[/tex]
Final speed [tex]\( v = 70.2 \, \text{m/s} \)[/tex]
Time [tex]\( t = 35.5 \, \text{s} \)[/tex]
Average acceleration a is given by:
[tex]\[ a = \frac{v - u}{t} \][/tex]
[tex]\[ a = \frac{70.2 \, \text{m/s} - 0 \, \text{m/s}}{35.5 \, \text{s}} = 1.41 \, \text{m/s}^2 \][/tex]
Complete question:
A 747 airliner reaches its takeoff speed of 157 mi/h in 35.5 s. What is the magnitude of its average acceleration?
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Rewritten by : Barada
The 747 airliner's average acceleration is found using the acceleration formula, which results in approximately 1.98 m/s². The conversion from miles per hour to meters per second was necessary for the calculation.
To find the magnitude of the airliner's average acceleration, we can use the formula for acceleration, which is:
a = (v - u) / t
where:
- v is the final velocity
- u is the initial velocity (0 in this case since the plane starts from rest)
- t is the time taken
First, we need to convert the takeoff speed from miles per hour to meters per second. Using the conversion factor (1 mile = 1609 meters, 1 hour = 3600 seconds):
157 mi/h × (1609 m / 1 mile) × (1 hour / 3600 s) ≈ 70.2 m/s
Now, we can plug these numbers into the formula:
a = (70.2 m/s - 0 m/s) / 35.5 s ≈ 1.98 m/s²
Thus, the magnitude of the average acceleration of the 747 airliner is approximately 1.98 m/s².
