Key Concepts: The gradient of a velocity-time graph gives acceleration, and the area under the graph gives distance.
Section A: Reading Graphs
1. What does a horizontal line on a velocity-time graph show? [2]
2. What does a line sloping upwards on a velocity-time graph show? [2]
Section B: Calculations
3. A car increases its velocity from $4\,m/s$ to $16\,m/s$ in 6 s. Calculate the acceleration. [2]
4. A train travels at $12\,m/s$ for 20 s. Calculate the distance traveled. [2]
5. A cyclist accelerates uniformly from $2\,m/s$ to $8\,m/s$ in 3 s. Calculate the acceleration and distance traveled in that time. [4]
6. A car travels at $10\,m/s$ for 5 s, then slows uniformly to rest over the next 5 s. Calculate the total distance traveled. [4]
Total marks: 16
Mark Scheme
1. Constant velocity [2]
2. Positive acceleration [2]
3. $a = (v - u)/t = (16 - 4)/6 = 2\,m/s^2$ [2]
4. $d = vt = 12 \times 20 = 240\,m$ [2]
5. $a = (8 - 2)/3 = 2\,m/s^2$; distance = area = $\tfrac{1}{2}(2 + 8) \times 3 = 15\,m$ [4]
6. Distance = rectangle $10 \times 5 = 50\,m$ plus triangle $\tfrac{1}{2} \times 10 \times 5 = 25\,m$; total $75\,m$ [4]