a. Velocity `= 55 × 3 = 165\ "ms"^-1`

b. The acceleration is a constant `55\ "ms"^-2`, so at `t = 3\ "s"`, the acceleration will be `55\ "ms"^-2`.

c. The distance travelled in `3` seconds is `165 × 1.5 = 247.5\ "m"`. We obtain this from the area under the line between `0` and `3` (i.e. the area of the shaded triangle below).

d. Note in the graph that we have **velocity** on the vertical axis, and the units are m/s.

The graph finishes at (3, 165).

Please support IntMath!