An object is moving back and forth on the x-axis according to the equation x(t) = 3sin(20πt), t> 0, where x(t) is measured in
1 answer:
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Since you've completed parts a and b, I will tackle part c.
For part C
To respond to this question, we must identify the zeros of the velocity function:
This polynomial can be factored:
Finding the zeros now becomes straightforward since the function equals zero when any factor is zero.
By solving these equations, we identify our zeros:
The particle remains stationary at t=0 and t=3/2.
For part D
We must discover when the velocity function exceeds zero. We will utilize its factored form.
We will assess when each factor is greater than zero and compile the findings in the following table:
From the table, it's evident that our function is positive when
and t>3/2.
This indicates the interval during which the particle moves forward.
For part E
The distance traveled can be represented as:
We simply substitute t=12 to calculate the total distance traveled:
For part F
Acceleration is defined as the rate at which velocity changes.
We determine acceleration by deriving the velocity function concerning time.
To find the acceleration at 1 second, we substitute t=1s into the previous equation:
For part C
To respond to this question, we must identify the zeros of the velocity function:
This polynomial can be factored:
Finding the zeros now becomes straightforward since the function equals zero when any factor is zero.
By solving these equations, we identify our zeros:
The particle remains stationary at t=0 and t=3/2.
For part D
We must discover when the velocity function exceeds zero. We will utilize its factored form.
We will assess when each factor is greater than zero and compile the findings in the following table:
From the table, it's evident that our function is positive when
This indicates the interval during which the particle moves forward.
For part E
The distance traveled can be represented as:
We simply substitute t=12 to calculate the total distance traveled:
For part F
Acceleration is defined as the rate at which velocity changes.
We determine acceleration by deriving the velocity function concerning time.
To find the acceleration at 1 second, we substitute t=1s into the previous equation:
Answer:
h = 10 m
Explanation:
Given,
mass of the platform = 50 Kg
Kinetic energy = 5000 J
height from which the diver dove =?
Taking the acceleration due to gravity as 10 m/s²
Using the conservation of energy principle
Kinetic energy is transformed into potential energy
K.E = P.E
K.E = m g h
5000 = 50 x 10 x h
500 h = 5000
h = 10 m
The height from which the diver dove is equal to h = 10 m