Topic: Dynamics
Concepts: Newton’s laws, Scale reading

---

Solution:

The scale reads the normal force $N$ and assumes it is the weight $mg$. The measured mass is then $m_{\text{scale}}=N/g$. If the elevator has acceleration $a$ upward, we have

so

In the beginning, the elevator is at rest so $a=0$ and $m_{\text{scale}}=m=80\,\mathrm{kg}$. Once the elevator starts moving, the acceleration can be found from the measured mass,

Thus, between $t=2\,\mathrm{s}$ and $4\,\mathrm{s}$ we have acceleration

Between $t=22\,\mathrm{s}$ and $24\,\mathrm{s}$ we have acceleration

The student has maximum downward velocity after the downward acceleration of the elevator but before it slows down with upward acceleration. This is from $t=4\,\mathrm{s}$ to $22\,\mathrm{s}$ so the answer is C.

Topic: Kinematics
Concepts: 5 kinematics equations, Motion graphs

---

Solution:

We found in the previous problem the acceleration of the elevator as a function of time:

where we take the positive direction to point downwards. To find the height of the building, we calculate the distance the elevator traveled from beginning to end. Between $t=2\,\mathrm{s}$ and $4\,\mathrm{s}$ ($\Delta t_1=4\,\mathrm{s}-2\,\mathrm{s}=2\,\mathrm{s}$), the elevator accelerates from rest so

At $t=4\,\mathrm{s}$, the elevator is moving with velocity $v=a\Delta t_1=(2.5\,\mathrm{m/s^2})(2\,\mathrm{s})=5\,\mathrm{m/s}$. From $t=4\,\mathrm{s}$ to $22\,\mathrm{s}$ ($\Delta t_2=22\,\mathrm{s}-4\,\mathrm{s}=18\,\mathrm{s}$), the elevator moves at constant velocity so

From $t=22\,\mathrm{s}$ to $t=24\,\mathrm{s}$ ($\Delta t_3=24\,\mathrm{s}-22\,\mathrm{s}=2\,\mathrm{s}$), the elevator decelerates to rest so

Thus,

so the answer is C.