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# Tutor profile: David L.

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David L.
Taught Mathematics and Physics for three years, BS Physics graduate, Master of Education (Mathematics) student
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## Questions

### Subject:Physics (Newtonian Mechanics)

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Question:

A student throws a ball vertically upward at an initial velocity of $$5\,\text{m/s}$$. Determine the velocity and the position of the ball at a time interval $$t = 1 \,\text{s}$$.

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David L.

We start off by listing down the formulas needed to solve the problem: $( v = v_0 - gt$) $( y = y_0 + v_0 t - \frac{1}{2} gt^2$) where $$v$$ is the velocity at some given time interval $$t$$, $$v_0$$ is the initial velocity, $$y$$ is the position at some given time interval $$t$$, $$y_0$$ is the initial position, and $$g$$ is the acceleration due to gravity, $$9.81\,\text{m/s}^2$$. First, we set up our coordinate system. Let the initial position of the ball be the origin (at the hand of the student), so we have $$y_0 = 0$$. Let the upwards direction be positive and the downwards direction be negative. Solving for the velocity, we have (noting that the initial velocity $$v_0 = 5 \, \text{m/s}$$): $( v = 5 - (9.81)(1)$) $( v = - 4.81$) $( v \approx - 5$) The velocity of the ball at $$t = 1 \,\text{s}$$ is $$-5 \, \text{m/s}$$ or $$5 \,\text{m/s}$$ downwards. Note that we rounded off to adhere to the proper use of significant figures. Solving for the position, we have: $( y = 0 + (5)(1) - \frac{1}{2} (9.81)(1)^2$) $( y = 10 - (9.81)(1)$) $( y = 0.19$) $( y \approx 0$) The position of the ball at $$t = 1 \,\text{s}$$ is $$0 \, \text{m}$$. Again, we rounded off the answer to adhere to the proper use of significant figures. At $$t = 1 \, \text{s}$$, the ball has approximately returned to its original position. That is why $$y \approx 0 \, \text{m}$$. We know that it has returned because its velocity is the same magnitude as its initial velocity ($$5 \, \text{m/s}$$), but is going in the opposite direction (downwards).

### Subject:Calculus

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Question:

Find the optimum value/s of the following function: $$f(x) = 2x^2 + 5x+3$$. Use differentiation in you answer.

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David L.

Note that the optimum of a function occurs at the point/s where the derivative is zero. Mathematically, the function $$f(x)$$ assumes optimum value/s at $$x$$ if: $( f'(x)=0$) So, we first find the derivative of the given function: $$f(x) = 2x^2 + 5x+3$$. Using the power rule for differentiation, we have: $( f'(x) = 4x + 5$) Next, we set this derivative equal to zero: $( f'(x) = 4x + 5 = 0$) Solving for $$x$$: $( 4x + 5 = 0$) $( 4x = -5$) $( x = -\frac{5}{4}$) This is the value of $$x$$ that gives the optimum value of the function. Substitute the value of $$x$$ to the original function: $( f\bigg(-\frac{5}{4}\bigg) = 2\bigg(-\frac{5}{4}\bigg)^2 +5\bigg(-\frac{5}{4}\bigg)+3$) Simplify the expression: $( f\bigg(-\frac{5}{4}\bigg) = 2\bigg(\frac{25}{16}\bigg) +5\bigg(-\frac{5}{4}\bigg)+3$) $( f\bigg(-\frac{5}{4}\bigg) = \bigg(\frac{25}{8}\bigg) -\bigg(\frac{25}{4}\bigg)+3$) $( f\bigg(-\frac{5}{4}\bigg) = \bigg(\frac{25}{8}\bigg) -\bigg(\frac{50}{8}\bigg)+\bigg(\frac{24}{8}\bigg)$) $( f\bigg(-\frac{5}{4}\bigg) = -\frac{1}{8}$) The function has only one optimum value. Its optimum value is $$-\frac{1}{8}$$.

### Subject:Algebra

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Question:

Find the roots of the following quadratic equation: $$3x^2 + 4x-5=0$$. Use the quadratic formula.

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David L.

Recall that for a quadratic equation with the following form: $$ax^2 + bx +c = 0$$, where $$a \neq 0$$, the quadratic equation is given as: $(x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$) In our case, the given quadratic equation is as follows: $$3x^2 + 4x - 5 = 0$$. We can, thus, identify $$a=3, b=4, c=-5$$. Using the formula outlined above, we have: $(x=\frac{-4 \pm \sqrt{(4)^2-4(3)(-5)}}{2(3)}$) Simplify the expression: $(x=\frac{-4 \pm \sqrt{16+60}}{6}$) $(x=\frac{-4 \pm \sqrt{76}}{6}$) $(x=\frac{-4 \pm 2\sqrt{19}}{6}$) Finally, we get: $(x=\frac{-2 \pm \sqrt{19}}{3}$) These are the roots we want.

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