(N/A) The distance covered by an object is equal to the area under the speed-time graph.
For car $P$,the area is a triangle with base $= 4\, s$ and height $= 6\, m/s$.
Distance $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4\, s \times 6\, m/s = 12\, m$.
For car $Q$,the area is a rectangle with length $= 4\, s$ and width $= 3\, m/s$.
Distance $= \text{length} \times \text{width} = 4\, s \times 3\, m/s = 12\, m$.
The difference in the distance travelled by the two cars is $12\, m - 12\, m = 0\, m$.
$(b)$ Yes,they move with the same speed at the point where the two graphs intersect. This occurs at $t = 2\, s$,where both cars have a speed of $3\, m/s$.
$(c)$ Car $P$ is undergoing uniform acceleration because its speed-time graph is a straight line passing through the origin,indicating a constant rate of change of speed. Car $Q$ is undergoing uniform motion (constant speed) because its speed-time graph is a horizontal line,indicating that its speed does not change with time.