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Velocity-Time Graph

2,064 bytes added, 13:16, 14 February 2019
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<math>area = distance = 160m</math>
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{| class="wikitable"
| style="height:20px; width:500px; text-align:left;" colspan = "3"|'''Calculate the [[distance travelled]] by the [[object]] at each stage in this journey.'''
|-
| style="height:20px; width:500px; text-align:left;" colspan = "3"|[[File:vtGraph1.png|centre|500px]]
|-
| style="height:20px; width:300px; text-align:left;" |'''A'''
| style="height:20px; width:300px; text-align:left;" |'''B'''
| style="height:20px; width:300px; text-align:left;" |'''C'''
|-
| style="height:20px; width:300px; text-align:left;" |'''State the known [[variable]]s.'''
[[File:vtGraphCalculateArea3.png|centre|150px]]
b = 2s
 
h = 30m/s
| style="height:20px; width:300px; text-align:left;" |'''State the known [[variable]]s.'''
[[File:vtGraphCalculateAcceleration3.png|centre|150px]]
b = 3s
 
h = 30m/s
| style="height:20px; width:300px; text-align:left;" |'''State the known [[variable]]s.'''
[[File:vtGraphCalculateAcceleration3.png|centre|150px]]
b = 3s
 
h<sub>1</sub> = 30m/s
 
h<sub>2</sub> = 50m/s
|-
| style="height:20px; width:200px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].'''
 
<math>area = \frac{b \times h}{2}</math>
 
<math>area = \frac{2 \times 30}{2}</math>
 
<math>area = \frac{60}{2}</math>
 
<math>area = distance = 30m</math>
| style="height:20px; width:200px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].'''
 
<math>area = b \times h</math>
 
<math>area = 3 \times 30</math>
 
<math>area = distance = 150m</math>
| style="height:20px; width:200px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].'''
 
<math>area = b \times h</math>
 
<math>area = 3 \times 30</math>
 
<math>area = distance = 90m</math> for lilac shaded area.
 
<math>area = \frac{b \times h}{2}</math>
 
<math>area = \frac{3 \times (50-30)}{2}</math>
 
<math>area = \frac{40}{2}</math>
 
<math>area = distance = 20m</math> for red area.
 
Total distance = 110m
|}