PHYS 2101 Final: HW1 Solutions
Document Summary
These solutions use the parameter values from the problems printed in the book, not those that appear in your personal homework assignment. The logic used to get to the answer is the same, however: the mass density of the candy is m. If we neglect the volume of the empty spaces between the candies, then the total mass of the candies in the container when filled to height h is. Thus, the rate of mass change is given by is the base area of the container that remains unchanged. (14. 0 cm)(17. 0 cm) 238 cm. A dh dt (4. 00 10 kg/cm )(238 cm )(0. 250 cm/s) 80 sm (b) the instantaneous velocity at time t is given by v = dx/dt = 50 + 20t, in. At t = 3. 0 s, v = 50 + (20)(3. 0) = 110 m/s. (c) the instantaneous acceleration at time t is given by a = dv/dt = 20 m/s2.
