79. If the takeoff velocity of an airplane on a runway is 300 km/hr with an acceleration of 2 m/s. What is the takeoff time of the airplane.

The take off time is equal to 41.66s.

Explanation:

Given that,

The initial speed of the airplane, u = 0

Acceleration, a = 2 m/s²

Let v is the take off time. Using first equation of motion,

v = u +at

Put v = 300 km/hr = 83.3 m/s

So,

So, the take off time is equal to 41.66s.

answer

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explanation:

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14