Question

6. Decelerating a plane at a uniform rate of -8 m/s2, a pilot stops the plane in 484 m. How

fast was the plane going before braking begins?

Answer 1

The plane was going approximately 88 m/s before braking begins, calculated using the kinematic equation for uniformly accelerated motion and the given deceleration rate and stopping distance.

Step-by-step explanation:

To find how fast the plane was going before braking begins, we can use the following kinematic equation for uniformly accelerated motion:

v^2 = u^2 + 2as

Where:

v is the final velocity, which is 0 m/s since the plane stops.

u is the initial velocity, which we want to find.

a is the deceleration, given as -8 m/s^2.

s is the stopping distance, which is 484 m.

Rearranging the equation to solve for u, we get:

u = √(v^2 - 2as)

Substituting the given values:

u = √(0 - 2(-8 m/s^2)(484 m))

u = √(2(8 m/s^2)(484 m))

u = √(7744 m^2/s2)

u ≈ 88 m/s

Therefore, the plane was going approximately 88 m/s before braking begins.

Answer 2

Answer: 88 m/s

Step-by-step explanation:

If we are talking about an acceleration at a uniform rate, we are dealing with constant acceleration, hence we can use the following equation:

`{V_(f)}^(2)={V_(o)}^(2)+2ad` (1)

Where:

`V_(f)=0` Is the final velocity of the plane (we know it is zero because we are told the pilot stops the plane at a specific distance)

`V_(o)` Is the initial velocity of the plane

`a=-8m/s^(2)` is the constant acceleration of the plane

`d=484m` is the distance at which the plane stops

Isolating
`V_(o)` from (1):

`V_(o)=√(-2ad)` (2)

`V_(o)=\sqrt{-2(-8m/s^(2))(484m)}` (3)

Finally:

`V_(o)=88m/s` This is the veocity the plane had before braking began