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.