Question

An aircraft starts at rest and is accelerated for 9.2 seconds, at which time the aircraft launches. If the distance traveled from the starting point to the launch point was 428.8 m, what was the launch velocity of the aircraft in m/s? Assume the acceleration is constant. (Hint: multi-step problem).

Answer 1

Explanation:

Since the acceleration is constant, we use the big four equations.

$x = x _{i} + vt$

$x = \frac{1}{2} a {t}^{2} + v _{i}t + x _{i}$

$v {}^{2} = (v _{i}) {}^{2} + 2a(x - x _{i})$

$\frac{1}{2} (v + v _{i})t = x - x _{i}$

Next. identity the variables.

We know

t=9.2

change in position or (x- x_i) is 428.8

Initial Velocity is 0.

We need to solve for the final velocity, so we use the fourth equation.

$\frac{1}{2} (v - v _{i})t = x - x _{i}$

Subsitue

$\frac{1}{2} (v - 0)(9.2) = 428.8$

$\frac{1}{2} (v)9.2 = 428.8$

$v = 93.22 \frac{m}{s}$