Answer from cs61c-ei (Jing Chen 16669246) for Question 2
The reason the sequence of instructions cannot run at full speed on the pipelined datapath without special hardware is because the add instruction depends on the value of $t0, which is obtained only at the very end of the lw instruction (data hazard). To solve this, we would need to not only forward the value of $t0 from the lw instruction to the add instruction but also stall. The reason for this is because the value of $t0 is available only after the fourth stage of the lw instruction, which is too late for the input of the third stage of add.