Question

4) Consider a separate rocket, also in deep space with a mass of 30.0 kg. If the rocket is

observed to be travelling at v= 5.00 m/s at t= 3.00 s and then travelling at û = -2.00 m/s at t= 14.0 s
with constant acceleration, calculate:
a) the acceleration, a.
b) the force F, acting on the rocket from the thrusters,

Answer 1

(a) -0.636 m/s²

(b) -19.08 N

Step-by-step explanation:

Given:

Mass of the rocket (m) = 30.0 kg

Initial velocity of rocket (v) = 5.00 m/s

Initial time of rocket (t₁) = 3.00 s

Final velocity of the rocket (u) = -2.00 m/s

Final time of rocket (t₂) = 14.0 s

(a)

Acceleration is given as the rate of change of velocity. Therefore,

`a=(u-v)/(t_2-t_1)\\\\a=(-2.00-5.00)/(14.00-3.00)\ m/s^2\\\\a=(-7.00)/(11.00)\ m/s^2\\\\a=-0.636\ m/s^2`

Therefore, the acceleration of the rocket is 0.636 m/s².

(b)

From Newton's second law, we know that, force acting on a body is equal to the product of its mass and acceleration.

So, the force acting on the rocket is given as:

`F=ma\\\\F=(30.0\ kg)(-0.636\ m/s^2)\\\\F=-19.08\ N`

Therefore, the force acting on the rocket is -19.08 N.

The negative sign implies the force acts in the direction opposite to motion.