How to find distance - CLEP Calculus

The t-shirt has no vertical velocity and initial height so its vertical position is given by the function

.

We can find the amount of time the t-shirt is airborn by setting .

We can make things a bit easier on ourselves by approximating

and then set that expression equal to zero.

This gives us seconds to reach the ground. Multiply this by the horizontal velocity to get the horizontal distance travelled, which is meters.

To find the intersection of the two circles, we will have to solve the simultaneous system:

.

We first write the second part of the system in a simplified manner , we have then:

and this gives:

.

Now we have :

, then this gives and hence:

and solving for y we have .

We use the fact that and therefore we have:

.

This gives us

.

Hence

.

Finally we deduce the value of y by plugging in each x value into our original equation and solving for y . When we do this we get , and when .

Now we the points and .

Using the distance formula we have :

The coordinates of the midpoint are simply the averages of the coordinates of the two points .

(\[-1+3\]/2, \[0+4\]/2, \[5+1\]/2)

=(2/2, 4/2, 6/2)

=(1,2,3)

To solve for this displacement, we can take the double integral of a(t). For simplicity, let us define things as follows:

v(t) = ∫(a(t) dt) and s(t) = ∫(v(t) dt)

The integrals are rather simple:

v(t) = sin(t) + 6t + c1

To solve for c1, we know that at t = 0, the initial velocity was 4. This means v(0) = sin(0) + 6 * 0 + c1 = 4; sin(0) = 0, so we know c1 = 4

Therefore, v(t) = sin(t) + 6t + 4

s(t) = –cos(t) + 3t2 + 4 * t + c2 (The final constant will drop out when we solve.)

Since we know that the body's position is always increasing (based on the initial positive velocity and an acceleration that must always be positive based on its equation), the total displacement is merely found by subtracting s(2) from s(4):

s(4) = –cos(4) + 3*42 + 16 + c2 = –cos(4) + 48 + 16 + c2 = –cos(4) + 64 + c2

s(2) = –cos(2) + 3*22 + 8 + c2 = –cos(2) + 12 + 8 + c2 = –cos(2) + 20 + c2

s(4) - s(2) = –cos(4) + 64 + cos(2) – 20 = 44.237496784317 (approx.) or 44.24

To find the change in position, we must take the definite integral of the velocity function

∫\[4, 8\](4t – 3)dt = \[4, 8\](2t2 – 3t) = (2 * 82 – 3 * 8) – (2 * 42 – 3 * 4)

= 2 * 64 – 24 – (2 * 16 – 12) = 104 – 20 = 84

To find the distance traveled by the object over a certain amount of time, we need an equation for its position. We are given an equation for its velocity, so if we integrate that equation from t=1 to t=2 seconds we'll obtain the distance traveled by the object over that interval:

In order to find the distance traveled by an object we need an equation for position. If velocity is the derivative of position, then we must integrate the given equation from t=2 to t=5 to find the total distance traveled by the object over that interval:

To find the midpoint between the two points, the average of the coordinates can be taken:

Where and

Plugging in these values we find our midpoint.

To find the displacement of the object, we can integrate the equation for acceleration twice.

Integrating the acceleration equation once will give:

Because the initial velocity = 2, we can set . Therefore,

Therefore, .

We can now use this in the velocity equation to give

integrating the velocity equation from to will give the distance

Let . We know that its symmetric with respect to the origin is given by:

b=.

Now we will use the distance formula to find the distance between a and b.

We have:

and this gives: .

This shows the required result.

To find the distance travelled, we must integrate the velocity equation

To find the distance between two points we need to find the difference of the x-coordinates and y-coordinates between the two points.

So the x-coordinate is

The y-coordinate is

So the x-coordinate and y-coordinate is

The distance can be calculated by

So plugging in

To find the change in position of the car, let us start with the car's acceleration as a function of time:

Since acceleration doesn't change with time, it has a constant value.

(the negative is used to represent deceleration).

We can then integrate this function with respect to time to find velocity.

Where v0 represents the initial velocity, which was given to us as 50 m/s.

We want to know the time where the car comes to rest, meaning where v(tr) = 0.

Solving

for tr gives our time at rest, 5 s.

Now, we can integrate our velocity function with respect to time to find our position function.

Since we're not interested in the absolute position at the time of rest, but rather the change in position, we can move the x0 term to the other side of the equation:

Plugging in our values for

and

We can find how far the car travelled after the brakes were hit, our

quantity, to be .

We will use the distance formula to show this result. Recall that having two points with coordinates , then the distance between this two points is given by:

. We will call the first fixed point . To find the second point, note that the equation of the circle with radius 1 is given by:

. Since we are looking for the upper half, we have by solving for y:

( We need only the upper half in this case , that is why y is .

Hence our second point is .

Using the distance formula we have:.

We need to simplify this expression a bit:

this gives finally after cancelling and adding:

We will use the distance formula to show this result. Recall that having two points with coordinates , then the distance between this two points is given by:

. We will call the first fixed point . To find the second point, note that the equation of the circle with radius 1 is given by:

.

Since we are looking for the lower half, we have by solving for y:

( We need only the lower half in this case , that is why y is .

Hence our second point is .

Using the distance formula we have:.

We need to simplify this expression a bit:

this gives finally after cancelling and adding:

We will use the distance formula to show this result. Recall that having two points with coordinates , then the distance between these two points is given by:

.

We will call the first fixed point .

To find the second point, note that the coordinate of the walking person is (x,4) since the x is changing and y is always 4.

Using the distance formula we have:.

We need to simplify this expression a bit:

this gives finally after cancelling and adding:

We will use the distance formula to show this result. Recall that having two points with coordinates , then the distance between these two points is given by:

.

We will call the first fixed point . To find the second point, note that the coordinates of the walking person is (x,0) since James is walking along the x-axis.

Using the distance formula we have:.

We need to simplify this expression a bit to get:

To find the center of the circle. We will have to rewrite the expression of the circle.

We have .

Using complete the square method, this gives:

and we know that:

Therefore we have :

is the same as : .

This means that the center is (1,1). Since the moving has (x,4) as coordinates (the y-coordinate is the same, x is changing).

Using the distance formula we have :

.

We can simplify this expression to get :

We will have to find these points. After that we will use the distance formula to find the distance between them. To find the intersection we will have to solve the following system:

which we can write by solving for y in the second equation.

Plugging in the value of y in the first system we have:

. This gives:

and hence we have:

.

This gives or .

We deduce the value of y from:

.

If .

If

We have the two points of interesection:. Now we use the distance formula to the distance :