a jetliner, traveling northward,is landing with a speed of 69m/s.once the plane touches down, it has a 750m runway in which speed is reduce to 6.1m/s. compute the avgerage acceleration (magnitude and direction) of the plane during landing.

vf^2=vi^2+2ad

solve for a

v^2=u^2 +2as

V=6.1
U=69
A=?
S=750
6.1^2=69^2 + 2 (a) (750)
37.21=4761+1500 a
37.21-4761/1500=a
a=-3.1m/s^2
^(raised to the power of or exponential)

To compute the average acceleration of the plane during landing, we need to use the formula:

average acceleration = (final velocity - initial velocity) / time

First, let's calculate the time it takes for the plane to come to a stop (i.e., from a speed of 69 m/s to 6.1 m/s).

Using the formula:

final velocity = initial velocity + (acceleration × time)

We can rearrange the equation to solve for time:

time = (final velocity - initial velocity) / acceleration

Given:
Initial velocity (u) = 69 m/s
Final velocity (v) = 6.1 m/s

Using the above equation, we can calculate the time it takes for the plane to come to a stop:

time = (6.1 m/s - 69 m/s) / acceleration

Next, we need to find the distance covered during this time. Assuming uniform acceleration, we can use the formula:

distance = (initial velocity + final velocity) × time / 2

Given:
Distance (d) = 750 m

We can rearrange the equation to solve for acceleration:

acceleration = (final velocity - initial velocity) × 2 / time

Now, we can substitute the known values and solve for acceleration:

1. Calculate the time:
time = (6.1 m/s - 69 m/s) / acceleration

2. Calculate the acceleration:
acceleration = (6.1 m/s - 69 m/s) × 2 / time

3. Substitute the distance equation and solve for acceleration:
750 m = (initial velocity + final velocity) × time / 2
750 m = (69 m/s + 6.1 m/s) × time / 2

Solving this equation will give us the value of time.

Finally, substitute the value of time into the acceleration equation to compute the average acceleration (magnitude and direction) of the plane during landing.

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