Answer:
The plane would need to travel at least
(
.)
The
runway should be sufficient.
Explanation:
Convert unit of the the take-off velocity of this plane to
:
.
Initial velocity of the plane:
.
Take-off velocity of the plane
.
Let
denote the distance that the plane travelled along the runway. Since acceleration is constant but unknown, make use of the SUVAT equation
.
Notice that this equation does not require the value of acceleration. Rather, this equation make use of the fact that the distance travelled (under constant acceleration) is equal to duration
times average velocity
.
The distance that the plane need to cover would be:
.
It is B: ionosphere. This has the ability to bounce radio signals
The equation of the car is given by the equation,
x(t) = 2.31 + 4.90t² - 0.10t⁶
If we are going to differentiate the equation in terms of x, we get the value for velocity.
dx/dt = 9.8t - 0.6t⁵
Calculate for the value of t when dx/dt = 0.
dx/dt = 0 = (9.8 - 0.6t⁴)(t)
The values of t from the equation is approximately equal to 0 and 2.
If we substitute these values to the equation for displacement,
(0) , x = 2.31 + 4.90(0²) - 0.1(0⁶) = 2.31
(2) , x = 2.31 + 4.90(2²) - 0.1(2⁶) = 15.51
Thus, the positions at the instants where velocity is zero are 2.31 and 15.51 meters.