Question

An airplane lands on a runway with a velocity of 150 m/s how far will it travel until it stops if its rate of deceleration is constant at -3 m/s^2?
what is the answer to this question? please answer it simply so i can understand
what is deceleration
how can i pass my AP physics B exam?
Final question is a rocket is propelled upward with an acceleration of 25m/s^2 for 5s after that time the engine is shut off, the rocket continues to move upward. the maximum height in meters that the rocket will reach is what?

Answer 1

Question #27844

An airplane lands on a runway with a velocity of 150 m/s how far will it travel until it stops if its rate of deceleration is constant at -3 m/s^2? what is the answer to this question? please answer it simply so i can understand what is deceleration how can i pass my AP physics B exam?

Final question is a rocket is propelled upward with an acceleration of 25m/s^2 for 5s after that time the engine is shut off, the rocket continues to move upward. the maximum height in meters that the rocket will reach is what?

Solution:

An equation of equally decelerated motion is:

$$S = v_0 t - \frac{1}{2} a t^2$$

were $v_0$ is the initial velocity

The deceleration is the negative acceleration, the vector of deceleration have direction opposite to motion.

Such as

$$v = v_0 - a t$$

the final velocity is equal to zero

$$v_0 = a t, \quad t = \frac{v_0}{a}$$ $$S = v_0 \frac{v_0}{a} - \frac{1}{2} a \left(\frac{v_0}{a}\right)^2 = \frac{v_0^2}{a} - \frac{1}{2} \frac{v_0^2}{a}$$ $$S = \frac{v_0^2}{a}$$ $$S = \frac{150^2}{3} = 7500 \, m$$

Answer 7500 m

As for question with rocket:

The maximum height is:

$$H = H_1 + H_2$$

where: $H_1$ is the height on the first stage of motion (equally accelerated motion)

$H_2$ is the distance on the next stage of motion – equally decelerated motion

$$H_1 = \frac{1}{2} a t^2$$

$H_{2} = \frac{v_{1}^{2}}{g}$ were g is the acceleration due the gravity $v_{1}$ is the velocity after first stage of motion.

$$v_{1} = a t$$ $$H_{2} = \frac{(a t)^{2}}{g}$$ $$H = \frac{1}{2} a t^{2} + \frac{a^{2} t^{2}}{g}$$ $$H = \frac{1}{2} 25 * 5^{2} + \frac{25^{2} * 5^{2}}{9.8} = 1906.89 \, \text{m}$$

Answer: 1906.89 m