$(a)$ Draw velocity-time graphs for the following cases:
$(i)$ When the object is at rest.
$(ii)$ When the object is thrown vertically upwards.
$(b)$ $A$ motorcyclist riding motorcycle $A$ who is travelling at $36 \, km \, h^{-1}$ applies the brakes and stops the motorcycle in $10 \, s$. Another motorcyclist of motorcycle $B$ who is travelling at $18 \, km \, h^{-1}$ applies the brakes and stops the motorcycle in $20 \, s$. Plot the speed-time graph for the two motorcycles. Which of the two motorcycles travelled farther before it came to a stop?

(N/A) For motorcycle $A$:
Initial velocity $u = 36 \, km \, h^{-1} = \frac{36 \times 5}{18} = 10 \, m \, s^{-1}$.
Final velocity $v = 0 \, m \, s^{-1}$.
Time $t = 10 \, s$.
For motorcycle $B$:
Initial velocity $u = 18 \, km \, h^{-1} = \frac{18 \times 5}{18} = 5 \, m \, s^{-1}$.
Final velocity $v = 0 \, m \, s^{-1}$.
Time $t = 20 \, s$.
Distance travelled is equal to the area under the speed-time graph.
Distance travelled by $A$ before stopping = Area of triangle $OPQ = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \, s \times 10 \, m \, s^{-1} = 50 \, m$.
Distance travelled by $B$ before stopping = Area of triangle $OMN = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \, s \times 5 \, m \, s^{-1} = 50 \, m$.
Conclusion: Both motorcycles travelled the same distance before they came to a stop.