Key Concepts: The gradient of a velocity-time graph gives acceleration ($a = \Delta v / \Delta t$). The area under a v-t graph gives distance. Equations of motion: $a = (v - u)/t$; $v^2 = u^2 + 2as$; $s = ut + \tfrac{1}{2}at^2$.
Section A — Reading Velocity-Time Graphs
1. Describe what each of the following lines on a velocity-time graph represents. [4]
a) A horizontal line
b) A straight line with positive gradient
c) A straight line with negative gradient
d) A line returning to zero
2. Explain how to find the distance travelled from a velocity-time graph. [2]
Section B — Acceleration Calculations
3. A car increases velocity from $4\,\text{m/s}$ to $16\,\text{m/s}$ in 6 s. Calculate the acceleration. [2]
4. A train slows from $30\,\text{m/s}$ to rest in 15 s. Calculate the deceleration. [2]
5. A car travels at $10\,\text{m/s}$ for 5 s, then slows uniformly to rest over the next 5 s. Calculate the total distance travelled. [3]
6. A cyclist accelerates uniformly from $2\,\text{m/s}$ to $8\,\text{m/s}$ in 3 s. Calculate the acceleration and distance travelled. [4]
Section C — SUVAT Equations
7. A car changes speed from $3\,\text{m/s}$ to $13\,\text{m/s}$ in 5 s. Calculate the acceleration. [2]
8. A runner starts from rest and accelerates at $2\,\text{m/s}^2$ over 20 m. Find the final speed using $v^2 = u^2 + 2as$. [3]
9. A cyclist slows from $12\,\text{m/s}$ to $4\,\text{m/s}$ over 16 m. Find the acceleration. [3]
10. A train accelerates at $1.5\,\text{m/s}^2$ for 6 s from rest. Find the final speed and distance travelled. [4]
11. A bike travels at $5\,\text{m/s}$ and accelerates at $1.0\,\text{m/s}^2$ for 8 s. Calculate the distance travelled using $s = ut + \tfrac{1}{2}at^2$. [3]
Total marks: 32
Mark Scheme
1. a) Constant velocity (no acceleration) [1]; b) Uniform acceleration (speed increasing at constant rate) [1]; c) Deceleration (speed decreasing at constant rate) [1]; d) Object comes to rest [1] [4]
2. Calculate the area under the graph [1]; area of rectangle = length × height; area of triangle = ½ × base × height [1] [2]
3. $a = (v - u)/t = (16 - 4)/6 = 12/6 = 2\,\text{m/s}^2$ [2]
4. $a = (0 - 30)/15 = -2\,\text{m/s}^2$; deceleration = $2\,\text{m/s}^2$ [2]
5. Rectangle: $10 \times 5 = 50\,\text{m}$ [1]; Triangle: $\tfrac{1}{2} \times 10 \times 5 = 25\,\text{m}$ [1]; Total = 75 m [1] [3]
6. $a = (8 - 2)/3 = 2\,\text{m/s}^2$ [2]; distance = area = $\tfrac{1}{2}(2 + 8) \times 3 = 15\,\text{m}$ [2] [4]
7. $a = (13 - 3)/5 = 2\,\text{m/s}^2$ [2]
8. $v^2 = 0 + 2 \times 2 \times 20 = 80$ [1]; $v = \sqrt{80} \approx 8.9\,\text{m/s}$ [2] [3]
9. $a = (v^2 - u^2) / 2s = (16 - 144)/32 = -128/32 = -4\,\text{m/s}^2$ [3]
10. $v = u + at = 0 + 1.5 \times 6 = 9\,\text{m/s}$ [2]; $s = \tfrac{1}{2}at^2 = 0.5 \times 1.5 \times 36 = 27\,\text{m}$ [2] [4]
11. $s = ut + \tfrac{1}{2}at^2 = 5 \times 8 + 0.5 \times 1.0 \times 64 = 40 + 32 = 72\,\text{m}$ [3]