a jet plane traveling at +87 m/s lands on a runway and comes to ret in 11 s.
a)Calculate its uniform acceleration
b)Calculate the distance it travels
0=v₀-at
a=v₀/t
s=at²/2
To solve this problem, we can use the equations of motion. The first equation we will use is:
v = u + at
where:
v = final velocity (0 m/s since the plane comes to rest)
u = initial velocity (87 m/s)
a = acceleration
t = time (11 s)
From the equation, we can rearrange to solve for acceleration:
a = (v - u) / t
a) To calculate the uniform acceleration:
First, we substitute the values into the equation:
a = (0 - 87) / 11
a = -87 / 11
a = -7.9 m/s²
So, the uniform acceleration of the jet plane is -7.9 m/s² (negative sign indicates deceleration).
b) To calculate the distance traveled by the jet plane:
We can use the second equation of motion:
s = ut + (1/2)at²
where:
s = distance
u = initial velocity (87 m/s)
t = time (11 s)
a = acceleration (-7.9 m/s²)
Substituting the values into the equation:
s = (87 * 11) + (1/2)(-7.9)(11²)
s = 957 - (1/2)(7.9)(121)
s = 957 - (604.9)
s = 352.1 meters
Therefore, the distance traveled by the jet plane is 352.1 meters.