Kinematics: Question 5

Worked solution

Step 1: Recall what the area under a velocity-time graph represents

On a velocity–time graph, the area between the line and the time axis gives the distance travelled (or, with sign, the displacement). Splitting the journey into its three described stages and finding the area of each is the most reliable method.

Step 2: Area of the first stage (uniform acceleration from rest)

For $0$ to $4.0\text{ s}$, the velocity rises uniformly from $0$ to $8.0\text{ m s}^{-1}$. This region is a triangle with base $4.0\text{ s}$ and height $8.0\text{ m s}^{-1}$: $s_1 = \tfrac{1}{2} \times 4.0 \times 8.0 = 16\text{ m}$

Step 3: Area of the second stage (constant velocity)

For the next $6.0\text{ s}$, the velocity is constant at $8.0\text{ m s}^{-1}$. This region is a rectangle: $s_2 = 6.0 \times 8.0 = 48\text{ m}$

Step 4: Area of the third stage (uniform deceleration to rest)

For the final $2.0\text{ s}$, the velocity falls uniformly from $8.0\text{ m s}^{-1}$ to $0$. This region is also a triangle, with base $2.0\text{ s}$ and height $8.0\text{ m s}^{-1}$: $s_3 = \tfrac{1}{2} \times 2.0 \times 8.0 = 8.0\text{ m}$

Step 5: Add the three areas

$s_{total} = s_1 + s_2 + s_3 = 16 + 48 + 8.0 = 72\text{ m}$

As a check, the total time accounted for is $4.0 + 6.0 + 2.0 = 12.0\text{ s}$, matching the full flight duration given in the question.

Why the other options are wrong

• A ($56\text{ m}$): this comes from leaving out the first triangular section entirely ($48+8=56$).
• B ($64\text{ m}$): this comes from leaving out the final triangular section entirely ($16+48=64$).
• D ($80\text{ m}$): this comes from treating the final deceleration phase as a rectangle instead of a triangle, i.e. forgetting the factor of $\tfrac{1}{2}$ ($16+48+16=80$).

Final answer

• Total distance travelled $= \boxed{72}\text{ m}$, option C.