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David C.

Math and Engineering Tutor

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Mechanical Engineering

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Question:

You are designing the runway for a local airport. The slowest plane at the airport can takeoff with a speed of 5 m/s^2. The fastest can takeoff with a speed of 10 m/s^2. All of the planes have the same takeoff speed of 65 m/s. What is the shortest length that you can design the runway?

David C.

Answer:

In order to answer this question you must apply one of the three equations of motion. Recall the three equations. (I) V = V_0 + A * theta(T) (II) X = X_0 + V_0 * theta(T) +.5A + theta(T)^2 (III) V^2 = V_0^2 +2A * (X - X_0) Based on the values given in the problem, we should use (iii). V^2 = V_0^2 +2A * (X - X_0). Step One: Plug the values into equation iii. 65^2 = 0^2 + 2 * 5 * theta(X) Step Two: Solve for theta(X) Final Answer: The runway must be at least 422.5 meters long.

Calculus

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Question:

A race car is traveling on a straight track at a velocity of 80 meters per second when the brakes are applied at time t=0 seconds. From the time t=0 to the moment the race car stops, the acceleration of the race car is given by a(t ) = -6t^2 - t meters per second per second. During this time period, how far does the race car travel?

David C.

Answer:

You are given the formula for acceleration. Recall the relationship between position, velocity and acceleration. (The derivative of position is velocity and the derivative of velocity is acceleration.) Step one: You must take the antiderivative of the given acceleration equation. The antiderivative of a(t)=-6t^2-t gives v(t)=-2t^3-.5t^2+c. Step two: Solve for the constant, c, in v(t)=-2t^3-.5t^2+c. . From the question statement you know when t=0 that v must equal 80. Plugging in these values into our equations yields that c must equal 80. Thus, v(t)=-2t^3-.5t^2+80. Step three: Solve for the time when v=0. So, solve this equation: 0=2t^3-.5t^2+80. This equation yields that t=3.33862 seconds. This is the amount of time that the race car travels before breaking for a stop. The problem is asking for distance so you must take the antiderivative once more in order to get the position equation. Step four: You must take the antiderivative of the found velocity equation. The antiderivative of v(t)=-2t^3-.5t^2+80 gives p(t)=-.5t^4-.167t^3+80t+c. Step five: Solve for the constant, c, in p(t)=-.5t^4-.167t^3+80t+c. From the question statement you know that t=0 p must also be 0.Plugging in these values yields that c=0, thus p(t)=-.5t^4-.167t^3+80t. Step six : You are ready to solve! Plug in the time found from step three into the position equation found in step five. This yields the final answer. Final Answer: The race car travels 198.77 meters before stopping.

Algebra

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Question:

Train A and Train B are moving towards each other in a collision course. They are 100 miles apart and will collide in 55 minutes. Train A is moving at 25 miles per hours. How fast is Train B moving in miles per hour?

David C.

Answer:

Known(s): Train A speed: 25 mph Distance apart: 100 miles Time till collision: 55 minutes Step 1: First you must find out how far Train A travels in the 55 minutes. Convert 25 'miles per hour' to 'miles per minute'. To do this, simply divide by 60 since there are 60 minutes in one hour. (25 / 60 = .4167 miles per minute) Step 2: Once you have the speed in miles per minute of Train A you can simply multiply that by the number of minutes it travels. (.4167 * 55 = 22.917 miles) Step 3: You now know that Train A travels 22.917 miles in 55 minutes. Since this is less than 100 miles Train B must travel the rest of the way. Solve for the distance Train B travels. (100 - 22.917 = 77.083 miles) Step 4: Now that you have the distance that Train B travels in the 55 minutes you can find out the speed by doing the reverse of steps 1 and 2. (77.083 miles / 55 minutes = 1.402 miles per minute) Step 5: Convert the 'miles per minute' found in Step 4 to 'miles per hour' because that is what the question ask for. (1.402 miles per minute * 60 minutes = 84.09 miles per hour) Final Answer: Train B is traveling 84 miles per hour

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