Question

A plane dives downward at an acceleration of 6.73 m/s2. If the plane’s initial speed was 350 km/h and its final speed was 400 km/h, how far did it travel in meters?

Answer 1

Let the initial speed be $v_0 = 350km/h = \dfrac{350}{3.6}m/s$, the final speed be $v_f = 400km/h = \dfrac{40}{3.6}m/s$., and the acceleration be $a = 6.73m/s^2$ . The time required for such increase in speed is:

$$t = \dfrac{v_f - v_0}{a}$$

During this time the plane travelled the following distance:

$$d =v_0t + \dfrac{at^2}{2} = v_0\dfrac{v_f - v_0}{a} + \dfrac{(v_f - v_0)^2}{2a}\\ \space \\ d = \dfrac{350}{3.6}\cdot \dfrac{\dfrac{400}{3.6}- \dfrac{350}{3.6}}{6.73} + \dfrac{(\dfrac{400}{3.6} - \dfrac{350}{3.6})^2}{2\cdot 6.73} \approx 214.97m$$

Answer. 214.97m.