AS Level Maths 9709: Velocity–Time Graphs. Gradient, Area and Average Velocity

A cyclist rides along a straight road for $40$ seconds, starting from rest at a fixed point $O$ . The velocity–time graph of her motion consists of three straight-line stages:

Stage 1 ( $0 \le t \le 10$ ): the velocity increases uniformly from $0$ to $8\text{ m s}^{-1}$ .
Stage 2 ( $10 \le t \le 30$ ): the velocity stays constant at $8\text{ m s}^{-1}$ .
Stage 3 ( $30 \le t \le 40$ ): the velocity decreases uniformly from $8\text{ m s}^{-1}$ back to $0$ .

(a) Find the acceleration of the cyclist during Stage 1 and during Stage 3. [2]

(b) By considering the area under each stage of the velocity–time graph, find the total distance travelled by the cyclist during the $40$ seconds. [4]

Part (a): Acceleration from the gradient of each stage

On a velocity–time graph, the gradient gives the acceleration.

Stage 1 ( $0\le t\le 10$ ): velocity changes from $0$ to $8\text{ m s}^{-1}$ over $10$ s.

$a_1 = \frac{8-0}{10-0} = \frac{8}{10} = 0.8\text{ m s}^{-2}$

Stage 3 ( $30\le t\le 40$ ): velocity changes from $8$ to $0\text{ m s}^{-1}$ over $10$ s.

$a_3 = \frac{0-8}{40-30} = \frac{-8}{10} = -0.8\text{ m s}^{-2}$

(Stage 2 has constant velocity, so its acceleration is $0$ , though this is not asked for here.)

Part (b): Total distance from the area under the graph

The area under a velocity–time graph gives the distance travelled. The graph is made of a triangle (Stage 1), a rectangle (Stage 2) and a triangle (Stage 3).

Stage 1 (triangle): base $=10$ s, height $=8\text{ m s}^{-1}$

$A_1 = \frac12 \times 10 \times 8 = 40\text{ m}$

Stage 2 (rectangle): base $=20$ s (from $t=10$ to $t=30$ ), height $=8\text{ m s}^{-1}$

$A_2 = 20 \times 8 = 160\text{ m}$

Stage 3 (triangle): base $=10$ s, height $=8\text{ m s}^{-1}$

$A_3 = \frac12 \times 10 \times 8 = 40\text{ m}$

Total distance:

$A_1+A_2+A_3 = 40+160+40 = 240\text{ m}$

Recompute independently as a check: $40+160=200$ , and $200+40=240$ . This matches, so the total distance is $240$ m.

Part (c): Average velocity

Average velocity is the total distance divided by the total time. Since the velocity is never negative throughout the journey (the cyclist never reverses direction), the total distance equals the total displacement, so this is a valid average velocity, not just average speed.

$\text{average velocity} = \frac{\text{total distance}}{\text{total time}} = \frac{240}{40} = 6\text{ m s}^{-1}$

Final answers

(a) Stage 1 acceleration $=\boxed{0.8\text{ m s}^{-2}}$ ; Stage 3 acceleration $=\boxed{-0.8\text{ m s}^{-2}}$
(b) Total distance $=\boxed{240\text{ m}}$
(c) Average velocity $=\boxed{6\text{ m s}^{-1}}$