A 747 jet, travelling at a speed of 70 m/s, touches down on a runway. The jet slows to rest at the rate of 2.0 m/s^2. Calculate the total distance the jet travels on the runway as it is brought to rest.

What does "to rest" mean? What velocity is "at rest"?

1200m

but qhats thhe Vf in the problem

Vf= final velocity

Vi= initial velocity

i knoww but how do u find out what tge final velocity is in this problem

To calculate the total distance the jet travels on the runway as it is brought to rest, we need to consider the initial speed, the deceleration rate, and the time it takes for the jet to come to a stop.

First, we can find the time it takes for the jet to come to a stop using the equation:

Final Velocity (Vf) = Initial Velocity (Vi) + (Acceleration (a) * Time (t))

Since the final velocity is 0 m/s (the jet comes to a stop), the equation becomes:

0 = 70 m/s + (-2.0 m/s^2) * t

Rearranging the equation, we get:

2.0 m/s^2 * t = 70 m/s

t = 70 m/s / 2.0 m/s^2
t = 35 seconds

Now, we can calculate the total distance using the equation of motion:

Distance (d) = Initial Velocity (Vi) * Time (t) + (1/2) * Acceleration (a) * Time^2 (t^2)

Plugging in the values:

d = 70 m/s * 35 s + (1/2) * (-2.0 m/s^2) * (35 s)^2

d = 2450 m - 1225 m

d = 1225 meters

Therefore, the total distance the jet travels on the runway as it is brought to rest is 1225 meters.

Vf^2=vi^2+2ad solve for d