Airplane takeoff Suppose that the distance an aircraft travels...

Question

Airplane takeoff Suppose that the distance an aircraft travels along a runway before takeoff is given by D = (10/9)t², where D is measured in meters from the starting point and t is measured in seconds from the time the brakes are released. The aircraft will become airborne when its speed reaches 200 km/h. How long will it take to become airborne, and what distance will it travel in that time?

Accepted Answer

Welcome back, everyone. A car accelerates from rest along a straight road, and its distance from the starting point is given by the equation S equals 5/4 T2, where S is the distance in meters and T is the time in seconds. If the car needs to reach a speed of 90 kilometers per hour to safely merge onto a highway, how long will it take to reach the speed and what distance will it cover? So first of all, as we understand the equation, as equals 54, T2d has distance measured in meters, right? And our velocity or speed is given in kilometers per hour. So what we're going to do is some convert 90 kilometers per hour into meters per second, and we're going to use dimensional analysis, meaning first of all, we're going to convert kilometers into meters. This means that we're multiplying by the fraction which has kilometers in the denominator. And meters and the numerator. And now let's add the relationship. We know that 1 kilometer is 1000 m, and then our next fraction is going to allow us to convert hours into seconds. So we want to cancel out hours, meaning we're including hours in the numerator and seconds in the denominator. And now let's provide the relationship. We know that 1 hour is equivalent to 3600 seconds. And now when we perform the calculation, we end up with 25 m per second. Now, we want to identify the velocity function. Which is the derivative of the position function. So we want to differentiate as of T. We can factor out the constant 5/4s and differentiate T squared, right? Applying the power rule, we got 54s multiplied by 2 T, which is 10 T divided by 2, and we can simplify this as 10T divided by 4. I'm sorry, right? And we can simplify this as 5T divided by 2. Now we know that at some point of time, let's call it T1, our velocity becomes 25 m per second, so we can say that 5T1. Divided by 2 is equal to 25, and now we can rearrange the equation and solve for T1. So T1 is going to be equal to 25 multiplied by 2 divided by 5. And this gives us 10 seconds, which is our first answer. So to reach. The speed of 90 kilometers per hour, we need 10 seconds. And then we are asked to identify the distance the car will cover, so we can go back into our position function, right? S of T and simply evaluate S at 0.10. So we're going to get 5/4s multiplied by 10 squad, which gives us 500 divided by 4. And that's 125 m, which is our second answer to this problem. Thank you for watching.

Key Concepts

Kinematics and Speed

Differentiation

Solving Quadratic Equations

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