## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

(a) The impulse exerted on the rocket by the engine is $15,000~N~s$ (b) The rocket reaches its maximum speed at the end of the 30.0-second period when the rocket exerts a force on the rocket. After the force ends, the rocket reaches its maximum speed of 110 m/s
(a) We can use the graph to find the impulse exerted on the object. The impulse is equal to the area under the force versus time graph. $J = \frac{1}{2}~F_{max}~t$ $J = \frac{1}{2}(1000~N)(30.0~s)$ $J = 15,000~N~s$ The impulse exerted on the rocket by the engine is $15,000~N~s$. (b) The rocket reaches its maximum speed at the end of the 30.0-second period when the rocket exerts a force on the rocket. We can use the impulse to find the final momentum $p_f$. $p_f = p_0+J$ $p_f = m~v_0+J$ $p_f = (425~kg)(75.0~m/s)+15,000~N~s$ $p_f = 46,875~N~s$ We can use the final momentum to find the velocity $v_f$ after the force ends. $m~v_f = p_f$ $v_f = \frac{p_f}{m}$ $v_f = \frac{46,875~N~s}{425~kg}$ $v_f = 110~m/s$ After the force ends, the rocket reaches its maximum speed of 110 m/s.