Answer:
(a) -0.636 m/s²
(b) -19.08 N
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,
[tex]a=\frac{u-v}{t_2-t_1}\\\\a=\frac{-2.00-5.00}{14.00-3.00}\ m/s^2\\\\a=\frac{-7.00}{11.00}\ m/s^2\\\\a=-0.636\ m/s^2[/tex]
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:
[tex]F=ma\\\\F=(30.0\ kg)(-0.636\ m/s^2)\\\\F=-19.08\ N[/tex]
Therefore, the force acting on the rocket is -19.08 N.
The negative sign implies the force acts in the direction opposite to motion.
