Question

Select the correct answer.
An airplane on a runway accelerates at 4.0 meters/second2 for 28.0 seconds before takeoff. How far does the plane travel on the runway before takeoff?

A.
1.6 × 103 meters
B.
1.7 × 103 meters
C.
3.1 × 103 meters
D.
3.3 × 103 meters

Answer 1

To find the distance traveled by the airplane on the runway before takeoff, we can use the formula for distance under constant acceleration, which is:

\[
d = v_i t + \frac{1}{2} a t^2
\]

where:
- \( d \) is the distance,
- \( v_i \) is the initial velocity (which is 0 m/s since the airplane starts from rest),
- \( a \) is the acceleration (4.0 m/s²),
- \( t \) is the time (28.0 seconds).

Since the initial velocity \( v_i \) is 0, we can simplify the formula:

\[
d = \frac{1}{2} a t^2
\]

Now we can plug in the values:

\[
d = \frac{1}{2} (4.0 \, \text{m/s}^2) (28.0 \, \text{s})^2
\]

Calculating \( (28.0)^2 \):

\[
(28.0)^2 = 784 \, \text{s}^2
\]

Now substitute back into the formula:

\[
d = \frac{1}{2} (4.0) (784)
\]
\[
d = 2.0 \times 784
\]
\[
d = 1568 \, \text{meters}
\]

Converting to scientific notation:

\[
d = 1.568 \times 10^3 \, \text{meters}
\]

Rounding gives approximately \( 1.6 \times 10^3 \, \text{meters} \).

Therefore, the correct answer is:

**A. 1.6 × 10³ meters**.