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?
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.
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?
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.
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?
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