## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) The impulse exerted on the rocket by the engine is $15,000~N~s$ (b) The rocket reaches its maximum speed of 110 m/s at the end of the 30.0-second period when the rocket exerts a force on the rocket.
(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.