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valkas [14]
2 years ago
7

You've always wondered about the acceleration of the elevators in the 101 story-tall Empire State Building. One day, while visit

ing New York, you take your bathroom scale into the elevator and stand on them. The scales read 140 lb as the door closes. The reading varies between 120 lb and 170 lb as the elevator travels 101 floors.i)What is the maximum acceleration upward?ii)What is the maximum magnitude of the acceleration downward?
Physics
1 answer:
love history [14]2 years ago
7 0

To develop this problem we will proceed to convert all units previously given to the international system for which we have to:

140 lb = 63.5 kg \rightarrow 63.5kg (9.8m/s) =622.3 N

120 lb = 54.4 kg \rightarrow 54.4kg (9.8m/s)= 533 N

170 lb = 77.1 kg \rightarrow 77.1 kg (9.8m/s) =756 N

PART A ) From the given values the minimum acceleration will be given for 120Lb and maximum acceleration when 170Lb is reached therefore:

F = 756 - 622.3

F = 133.7N

Through the Newtonian relationship of the Force we have to:

F= ma

a = \frac{F}{m}

a = \frac{133.7}{63.5}

a = 2.1m/s^2

PART B) For the maximum magnitude of the acceleration downward we have that:

F = 622.3 - 533

F = 89.3N

Through the Newtonian relationship of the Force we have to:

F= ma

a = \frac{F}{m}

a = \frac{89.3}{63.5}

a = 2.1m/s^2

a = 1.04 m/s^2

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The position of a particle moving along the x-axis varies with time according to x(t) = 5.0t^2 − 4.0t^3 m. Find (a) the velocity
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<h2>Answer:</h2>

(a) v(t) = [10.0t - 12.0t²] m/s  and a(t) = [10.0 - 24.0t ] m/s² respectively

(b) -28.0m/s and -38.0m/s² respectively

(c) 0.83s

(d) 0.83s

(e) x(t)  = 1.1573 m           [where t = 0.83s]

<h2>Explanation:</h2>

The position equation is given by;

x(t) = 5.0t² - 4.0t³ m           --------------------(i)

(a) Since velocity is the time rate of change of position, the velocity, v(t), of the particle as a function of time is calculated by finding the derivative of equation (i) as follows;

v(t) = dx(t) / dt = \frac{dx}{dt} = \frac{d}{dt} [ 5.0t² - 4.0t³ ]

v(t) = 10.0t - 12.0t²     --------------------------------(ii)

Therefore, the velocity as a function of time is v(t) = 10.0t - 12.0t² m/s

Also, since acceleration is the time rate of change of velocity, the acceleration, a(t), of the particle as a function of time is calculated by finding the derivative of equation (ii) as follows;

a(t) = dx(t) / dt = \frac{dv}{dt} =  \frac{d}{dt} [ 10.0t - 12.0t² ]

a(t) = 10.0 - 24.0t             --------------------------------(iii)

Therefore, the acceleration as a function of time is a(t) = 10.0 - 24.0t m/s²

(b) To calculate the velocity at time t = 2.0s, substitute the value of t = 2.0 into equation (ii) as follows;

=> v(t) =  10.0t - 12.0t²

=> v(2.0) = 10.0(2) - 12.0(2)²

=> v(2.0) = 20.0 - 48.0

=> v(2.0) = -28.0m/s

Also, to calculate the acceleration at time t = 2.0s, substitute the value of t = 2.0 into equation (iii) as follows;

=> a(t) = 10.0 - 24.0t

=> a(2.0) = 10.0 - 24.0(2)

=> a(2.0) = 10.0 - 48.0

=> a(2.0) = -38.0 m/s²

Therefore, the velocity and acceleration at t = 2.0s are respectively -28.0m/s and -38.0m/s²

(c) The time at which the position is maximum is the time at which there is no change in position or the change in position is zero. i.e dx / dt = 0. It also means the time at which the velocity is zero. (since velocity is dx / dt)

Therefore, substitute v = 0 into equation (ii) and solve for t as follows;

=> v(t) = 10.0t - 12.0t²

=> 0 = 10.0t - 12.0t²

=> 0 = ( 10.0 - 12.0t ) t

=> t = 0            or             10.0 - 12.0t = 0

=> t = 0            or             10.0 = 12.0t

=> t = 0            or             t = 10.0 / 12.0

=> t = 0            or             t = 0.83s

At t=0 or t = 0.83s, the position of the particle will be maximum.

To get the more correct answer, substitute t = 0 and t = 0.83 into equation (i) as follows;

<em>Substitute t = 0 into equation (i)</em>

x(t) = 5.0(0)² - 4.0(0)³ = 0

At t = 0; x = 0

<em>Substitute t = 0.83s into equation (i)</em>

x(t) = 5.0(0.83)² - 4.0(0.83)³

x(t) = 5.0(0.6889) - 4.0(0.5718)

x(t) = 3.4445 - 2.2872

x(t)  = 1.1573 m

At t = 0.83; x = 1.1573 m

Therefore, since the value of x at t = 0.83s is 1.1573m is greater than the value of x at t = 0 which is 0m, then the time at which the position is at maximum is 0.83s

(d) The velocity will be zero when the position is maximum. That means that, it will take the same time calculated in (c) above for the velocity to be zero. i.e t = 0.83s

(e) The maximum position function is found when t = 0.83s as shown in (c) above;

Substitute t = 0.83s into equation (i)

x(t) = 5.0(0.83)² - 4.0(0.83)³

x(t) = 5.0(0.6889) - 4.0(0.5718)

x(t) = 3.4445 - 2.2872

x(t)  = 1.1573 m            [where t = 0.83s]

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