## Physics: Principles with Applications (7th Edition)

Changing the velocities to meters per second: $v_1 = (35~km/h)(\frac{1000~m}{1~km})(\frac{1~h}{3600~s}) = 9.7~m/s$ $v_2 = (65~km/h)(\frac{1000~m}{1~km})(\frac{1~h}{3600~s}) = 18.1~m/s$ Now, we find the power: $P = \frac{\Delta E}{\Delta t} = \frac{\frac{1}{2}(m)(v_2^2-v_1^2)}{t}$ $P = \frac{\frac{1}{2}(1300~kg)((18.1~m/s)^2-(9.7~m/s)^2)}{3.8~s}$ $P = 39,940~W$ We can use this power to find the time to accelerate from 55 km/h to 95 km/h: $v_1 = (55~km/h)(\frac{1000~m}{1~km})(\frac{1~h}{3600~s}) = 15.3~m/s$ $v_2 = (95~km/h)(\frac{1000~m}{1~km})(\frac{1~h}{3600~s}) = 26.4~m/s$ $P = \frac{\Delta E}{\Delta t} = \frac{\frac{1}{2}(m)(v_2^2-v_1^2)}{t}$ $t = \frac{\frac{1}{2}(1300~kg)((26.4~m/s)^2-(15.3~m/s)^2)}{39,940~W}$ $t = 7.5~s$ It would take 7.5 seconds to accelerate from 55 km/h to 95 km/h.