Kinematics in One Dimension: Car Example
Set of examples in kinematics focusing on a car accelerating uniformly, decelerating, and calculating total distance travelled. Explains concepts through graphs, calculations, and extended thoughts.
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Presentation Transcript
Kinematics motion in one dimension These are a set of examples, with each focusing on a different concept in this topic.
Worked Example 1 An introduction A car starts from rest and accelerates uniformly at a rate of 3 m/s for 10 seconds. It travels at this speed for 8 seconds before coming to rest 5 seconds later. i. Represent the information on a time-velocity graph. ii. Find the magnitude of the deceleration. iii. Find the total distance travelled.
Below is how a time-velocity graph of the motion looks .
To find the deceleration, we should consider what information we have about the scenario, and then choose an appropriate formula: u = 30m/s v= 0m/s t = 5s Using v = u + at, we can establish: 0 = 30 + a(5) -30 = 5a -6 = a ..note a negative acceleration indicates deceleration. So, the deceleration is 6m/s2
The total distance can be found by summing/adding the areas of the three shapes:
Using this method gives us: Total Area = Area Shape1+ Area Shape2+ Area Shape3 Total Area = 0.5(10)(30) + 8(30) + 0.5(5)(30) Total Area = 150 + 240 + 75 Total Area = 465 m2
Some extended thoughts. a) The ratio of the acceleration and deceleration to the times spent accelerating/decelerating are inversely proportional the length of time accelerating from 0 to 30 is twice the time spent decelerating from 30 to 0. Hence, the magnitude of the deceleration is twice that of the acceleration. This concept is examined frequently in more challenging questions.
