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# Tutor profile: Derek S.

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Derek S.
Market Analyst, beginner tutor
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## Questions

### Subject:Basic Math

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Question:

What is the perimeter and area of a rectangle with sides equal to 5 inches and 7 inches?

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Derek S.

The area of a rectangle is the base times the width = 5x7=35 sq. inches The perimeter of a rectangle is the sum of the length of all sides. 5+7+5+7=24 inches

### Subject:Calculus

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Question:

A hot air balloon is rising straight up from a level field. You observe this from 100 m away. When the angle from the liftoff point to the balloon reaches: $$\frac{\pi}{6}$$ the angle is increasing at a rate of 0.2 rad/min. How fast is the balloon rising at this moment?

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Derek S.

Since we are looking for $$\frac{dy}{dt}$$ and the only reference to this vertical velocity is the change in angle at the observer ($$\theta$$), we can solve for $$y$$ in terms of this angle and use the chain rule to differentiate with respect to $$t$$. $(tan(\theta) = \frac{y}{100}$) $(y=100* tan (\theta)$) Now using the chain rule: $(\frac{dy}{dt}=100*\sec^2(\theta) \frac{d\theta}{dt}$) and $(\frac{d\theta}{dt}=.2\ rad/s\ and\ \theta=\pi/6$) $(\frac{dy}{dt}=100*.2* \sec^2(\pi/6)=26.67 m/s$)

### Subject:Physics

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Question:

Assume a vacuum (i.e. no friction). All vectors along the x-axis or 90 degrees.: A force F of 20N is applied to a body B with mass of 5 kg at rest. A.) If F is applied for 3 seconds, what is the velocity V of B, after the 3 seconds? B.) What if F is applied for 5 seconds? C.) What if F is 40N for 2 seconds?

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Derek S.

Newton's second law: F=ma. a.) F=20N m=5kg so a=4 m/s^2 Since a is the derivative of velocity, integrate a from t=0 to t=3 V(t)=4t v=4(3)-4(0)=12 m/s b.) same as above from t=0 to t=5 V(t)=4t v=4(5)-4(0)= 20 m/s c.) F=40N m=5kg so a=8 m/s^2 V(t)=8t Integrate from t=0 to t=2 v=8(2)-8(0) = 16 m/s

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