A jet fighter lands on an aircraft carrier. m = 35,000 kg, the force of deceleration is 43,000 N, and the initial velocity is 27 m/s. What distance does it travel during the first 7.5 seconds of deceleration?
F=ma,
a=F/m,
s=v^2/2•a=m•v^2/2•F=
=35000•27^2/2•43000=296.7 m
To find the distance traveled during the first 7.5 seconds of deceleration, we can use the equation of motion:
d = (vi * t) + (1/2) * a * t^2
Where:
d = distance traveled during deceleration
vi = initial velocity
t = time of deceleration
a = acceleration (in this case, it is the negative of the deceleration)
Given:
vi = 27 m/s
t = 7.5 s
a = -43000 N
First, we need to convert the force of deceleration from Newtons to acceleration in m/s^2.
To do this, we can use Newton's second law:
F = m * a
Rearranging the equation, we can solve for acceleration:
a = F / m
Substituting the given values:
a = -43000 N / 35000 kg
Now, we can substitute the values of vi, t, and a into the equation of motion to find the distance traveled:
d = (27 m/s * 7.5 s) + (1/2) * (-43000 N / 35000 kg) * (7.5 s)^2
Calculating this expression will give us the distance traveled during the first 7.5 seconds of deceleration.