Module 2 discussion -find it!.docx - Module 2 discussion find it I think I have misunderstood the equation The correct one is given below Given values

Module 2 discussion -find it!.docx - Module 2 discussion...

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Module 2 discussion- find it! I think I have misunderstood the equation. The correct one is given below. Given values for a and b are, a= 2 and b= 7 s=4[a]t−4.[b]t2 Substitute a and b with the given values, s=42t−4.7t 2 ds/dt=42d(t)/dt−4.7d(t 2 )/dt ds/dt=v v=42−4.7×2×t v=42−9.4t 1. Find t for v=0ms -1 v=42−9.4t Substitute v=0, 0=42−9.4t 9.4t=42 t=429.4
4. What conclusions can you draw?
- When t=4s, the velocity has a positive value that indicates the object is still moving upwards (at t=4).- When t=5s, the velocity has a negative value that indicates the objects is moving in theopposite direction (downwards) from the direction that is considered as positive (in this problem the positive direction is upwards).-The object moves upwards and reaches the maximum height when t=4.468s (v=0ms-1) and then it starts to move to the opposite direction (downwards) after t=4.468s.

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