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# Tutor profile: Joe V.

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Joe V.
Senior Control System Engineer at Flowserve Corporation
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## Questions

### Subject:Physics (Newtonian Mechanics)

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Question:

An object with a mass of m = 10 kg, is dropped from a height of h = 50 meters, with no initial velocity. Assume a constant gravitational acceleration of g = 9.81 m/(s^2) during the object's fall. Ignoring the effects of drag, what is the speed of the object just before it hits the ground? Does the speed change if the mass is doubled?

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Joe V.

The key word here is "constant acceleration". You can use the constant acceleration kinematics equations to find the answer. start by listing your known variables: m = 10 kg h = 50 m v_initial = 0 m/s g = 9.81 m/(s^2); NOTE! it's important to define a sign convention for acceleration. For this example we will assume the downward acceleration as having a negative sign identify the kinematic equation best used, given the known variables. The kinematic equation best to use is: a (y_final - y_initial) = ( (v_final)^2 ) - ( (v_initial)^2 ) the variable of interest is "v_final" as mentioned in the "NOTE" above, we are assuming the acceleration is negative. Therefore: a = - g = - 9.81 (y_final - y_initial) Since we have defined the downward acceleration as negative, it makes sense that downward position be negative as well. We will define the y_initial as the origin or zero, and since the object is now "below" the origin, we will define y_final = - 50. Therefore: ( y_final ) - ( y_initial ) = ( -50 ) - ( 0 ) = - 50 now plug in the known variable to the select kinematics equation and solve for v_final. a (y_final - y_initial) = ( (v_final)^2 ) - ( (v_initial)^2 ) (- 9.81) ( -50 ) = ( (v_final)^2 ) - ( 0 ) 490.5 = ( (v_final)^2 ) taking the square root of both sides: 22.14 = v_final Notice the kinematic equations are independent of mass (m is not a variable). Therefore, an object with double the mass will have the same speed if dropped from 50 m.

### Subject:Trigonometry

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Question:

Suppose a right triangle has following lengths: | | y = 4 | |_ _ _ _ x = 3 Find the length of the hypotenuse

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Joe V.

Apply Pythagorean's Theorem: (x^2) + (y^2) = (h^2) where h = hypotenuse. plugging in the value of x and y: (3^2) + (4^2) = (h^2) 9 + 16 = (h^2) 25 = (h^2) taking the square root of both sides, yields: 5 = h (hint: a 3-4-5 triangle is a good one to remember off hand, as it is a commonly used in engineering and mathematics problems)

### Subject:Calculus

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Question:

Suppose an object's position x(t), at any given time (t), can be defined by the function: x(t) = 4(t^2) - 16t + 3; Does the object change its direction between t = 0 to t = 10? If so, at what time, "t", does it change its direction?

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Joe V.

One method of finding if and when the object changes its direction is to find the derivative of the function, x(t). Remember, the importance of a derivative - it is a function that describes the "instantaneous" slope at every point on x(t). For an object to change its direction, there must be a point where the derivative or the slope equals zero. The derivative of x(t): x'(t) = 8t - 16 other notation: dx x'(t) = ------- dt Setting x'(t) = 0 and finding the value of t, yields: 0 = 8t - 16 factoring for simplicity: 0 = 8 (t - 2) t = 2 The next step is to verify that t = 2 is in fact a point when the object changes its direction. To do this, pick a point before and after the time, t = 2 and plug it into the derivative function, x'(t), then compare not the value, but the sign. arbitrarily picking t = 1 and plugging into the derivative function x'(1) = 8(1) - 16 x'(1) = - 8; negative sign arbitrarily picking t = 3 and plugging into the derivative function x'(3) = 8(3) - 16 x'(3) = 24 - 16 x'(3) = + 8; positive sign Notice the sign changes from negative to positive before and after time t = 2, respectively. This means that the object does in fact change position at t = 2. Had the signs of the value remained the same, this would reveal the object "paused", but continued to move in the same direction. Convince yourself by plotting the original function, x(t) around t = 2.

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