A plane cruising at 247 m/s accelerates at

28 m/s
2
for 8.7 s.How far will it have traveled in that time?

s = vt + 1/2 at^2

Now just plug in v=247, a=28, t=8.7

3208.56

To find the distance traveled by the plane, we can use the formula:

Distance = Initial velocity x Time + (1/2) x Acceleration x Time^2

First, let's identify the given values:

Initial velocity (u) = 247 m/s
Acceleration (a) = 28 m/s^2
Time (t) = 8.7 s

Now, substitute the values into the formula:

Distance = (247 m/s) x (8.7 s) + (1/2) x (28 m/s^2) x (8.7 s)^2

Simplifying the equation:

Distance = 2150.9 m + (1/2) x 28 x (75.69 s^2)

Distance = 2150.9 m + (1/2) x 28 x 5739.24 s^2

Distance = 2150.9 m + 80012.96 m

Distance = 82163.86 m

Therefore, the plane will have traveled approximately 82163.86 meters (or 82.16 kilometers) in that time.