An airplane takes off at t=0 hours flying due north. It...

An airplane takes off at $t=0$ hours flying due north. It takes 24 minutes for the plane to reach cruising altitude, and during this time its ground speed (or horizontal velocity), in $\mathrm{mph}$, is $$\frac{d x}{d t}=1250-\frac{1050}{t+1}$$ and its vertical velocity, in $\mathrm{mph}$, is $$\frac{d y}{d t}=-625 t^{2}+250 t$$ (a) What is the cruising altitude of the plane? (b) What is the ground distance (or horizontal distance) covered from take off until cruising altitude? (c) Find the total distance traveled by the plane from take off until cruising altitude.

An airplane takes off at $t=0$ hours flying due north. It takes 24 minutes for the plane to reach cruising altitude, and during this time its ground speed (or horizontal velocity), in $\mathrm{mph}$, is
$$\frac{d x}{d t}=1250-\frac{1050}{t+1}$$
and its vertical velocity, in $\mathrm{mph}$, is
$$\frac{d y}{d t}=-625 t^{2}+250 t$$
(a) What is the cruising altitude of the plane?
(b) What is the ground distance (or horizontal distance) covered from take off until cruising altitude?
(c) Find the total distance traveled by the plane from take off until cruising altitude.

Key Concepts

Definite Integration

Definite integration is the process of finding the accumulated quantity, such as displacement or altitude, over a specific time interval by integrating a rate function. In problems involving velocity, integrating the velocity function over time yields the total displacement or change in position, which is a key method in kinematics.

Parametric Equations of Motion

Parametric equations represent motion in terms of functions of time for each coordinate, allowing separate analysis of horizontal and vertical components. This approach is essential when dealing with multi-dimensional motion, as it enables the computation of displacement components by integrating the respective velocity functions.

Arc Length of a Curve

Arc length calculation involves determining the total distance traveled along a curved path by integrating the speed, which is the magnitude of the velocity vector. This is done using the integral of the square root of the sum of the squares of the horizontal and vertical velocity components over time, providing the complete path length rather than just the net change in position.

Transcript

00:01 Dx upon dt equal to 1 ,250 negative 1 ,050 upon t plus 1.

00:14 It is horizontal velocity and vertical velocity is dy upon dt equal to negative 625 times t2.

00:31 T square plus 250 times t so here an aeroplane tax off at t equal to zero or flying due north it tags 24 minutes for plane to reach cruising altitude and during this time it is ground speed dx up on d t and d y up dd so here for first what is the cruising altitude of the plane so here given t equal to 24 minute so we have to convert minute to or so here we have to divide by 60 so here 24 time 1 upon 60 equal to 1 to 0 .4 are the integral of velocity is the net distance travel and we can just integrate d y upon d t since we only want to the vertical distance so here at time t equal to 0 to 0 .4 are these are the limits for integration.

02:10 Now, integral definite integral of negative 625 times t squared plus 250 time t d t from 0 to 0 .4.

02:31 So, now we have to find the anti -derivative of negative 625 times t squared plus 250 time t is negative 625 upon 3 times t cubed plus 125 times t squared from 0 to 0 .4.

03:02 So now here we have to put upper value which is 0 .4 so here t equal to 0 .4 so you put here negative 625 upon 3 time 0 .4 to the power 3 plus 125 time 0 .4 to the power 3 plus 125 time 0 .4 to the power 2 2.

03:35 Now here you can see that lower value is 0.

03:38 So here we get 0 when we put t equal to 0.

03:45 Now we simplify this and here we get approximate value 6 .67 miles.

03:56 So it is our final answer for part a.

03:59 Now we solve part b.

04:01 So here what is the ground distance covered from 10? check off until cruising altitude.

04:13 So here we know that for ground distance, distance will be horizontal.

04:19 So at t equal to 0 to t equal to 0 .4, and here we use d x upon d t.

04:35 So here definite integral of 1 ,250 negative 150 upon t plus 1 plus 1 time d t from 0 to 0 .4.

04:56 So now here we have to find 1 ,250 time 1 ,250 time, t negative 150 times natural modulus t plus 1 .4.

05:26 So now here you can see that upper value is 0 .4...