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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= 7s=4[a]t−4.[b]t2Substitute a and b with the given values,s=42t−4.7t2ds/dt=42d(t)/dt−4.7d(t2)/dtds/dt=v∴v=42−4.7×2×tv=42−9.4t1. Find t for v=0ms-1v=42−9.4tSubstitute v=0,0=42−9.4t9.4t=42t=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.