We apply Eq. [7 -4] twice: once for the man and once for the cheetah. We'll take the initial time t1 = 0, and at this time choose the initial position of the cheetah to be x1(cheetah) = 0, so that x1(man) = 0.400 km. We then solve for the time t 2 When x2(cheetah) = x 2(man), i.e., (110 km/h)at2 = 0.400km + (35.0km/h)1ff2 Therefore, 0400 km t = =5.33 >