Distance - AP Calculus AB

Card 0 of 1652

Question

A t-shirt is fired from a t-shirt cannon positioned high above a crowd at a concert. The t-shirt is fired horizontally with a velocity of .

How many meters does the t-shirt travel from the base of the t-shirt cannon before it hits the ground?

Answer

You cannot determine this without knowing how high the cannon is when it fires the t-shirt.

Compare your answer with the correct one above

Question

A t-shirt is fired from a t-shirt cannon positioned high above a crowd at a concert. The t-shirt is fired horizontally with a velocity of .

Supposing the cannon is above the ground. How many meters does the t-shirt travel laterally before hitting the ground?

Answer

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

$y(t)=-\frac{9.8}{2}t^2+0t+10$ .

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

$y(t)\approx -5t^2+10$ and then set that expression equal to zero.

This gives us $t=\sqrt2$ seconds to reach the ground. Multiply this by the horizontal velocity to get the horizontal distance travelled, which is $15\sqrt2$ meters.

Compare your answer with the correct one above

Question

We want to measure the distance of the intersection of the two circles:

and . What is this distance?

Answer

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

$\left{\begin{matrix} x^2+y^2=1\ (x-1)^2+(y-1)^2=1 \end{matrix}\right.$ .

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

$x=0\quad x=1$ .

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 :

$\sqrt{(1-0)^2+(1-0)^2}=\sqrt{2}$

Compare your answer with the correct one above

Question

Find the midpoint of the line segment connecting the points (-1,0,5) and (3,4,1).

Answer

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)

Compare your answer with the correct one above

Question

If the acceleration of a body at time t is given by a(t) = cos(t) + 6, what is the displacement of this particle from time t = 2 to time t = 4 if its initial velocity was 4 m / s?

Answer

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) + 342 + 16 + c2 = –cos(4) + 48 + 16 + c2 = –cos(4) + 64 + c2

s(2) = –cos(2) + 322 + 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

Compare your answer with the correct one above

Question

The velocity of a given particle at time t is given by the equation v(t) = 4t – 3

How far did the particle travel from time t = 4 to time t = 8?

Answer

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

Compare your answer with the correct one above

Question

The velocity of an object, in meters per second, is given by the following equation:

Find the distance traveled by the object from to seconds.

Answer

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:

$\int_{1}^{2}v(t)dt=\int_{1}^{2}(30t+7)dt=[15t^2+7t]_1^2$

Compare your answer with the correct one above

Question

Find the distance traveled by an object from to seconds if the velocity of the object, in m/s, is described by the following equation:

$v(t)=\frac{1}{4}t-\frac{1}{2}$

Answer

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:

$\int_{2}^{5}\left(\frac{1}{4}t-\frac{1}{2}\right)dt=\left[\frac{1}{8}t^2-\frac{1}{2}t\right]_{2}^{5}=\left(\frac{25}{8}-\frac{5}{2}\right)-\left(\frac{1}{2}-1\right)=\frac{9}{8}$

Compare your answer with the correct one above

Question

Find the midpoint of the line segment connecting the points and .

Answer

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

$\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right )$

Where and

Plugging in these values we find our midpoint.

$\left(\frac{-2+6}{2},\frac{7-3}{2}\right) = \left(\frac{4}{2},\frac{4}{2}\right) = (2,2)$

Compare your answer with the correct one above

Question

If the acceleration of an object at time is given by , what is the displacement of this particle from to , if its initial velocity was .

Answer

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

Integrating the acceleration equation once will give:

$v(t) = \frac{t^{2+1}}{2+1}+\frac{3t^{1+1}}{1+1}+12\cdott+C = \frac{1}{3}t^3 + \frac{3}{2}t^2+12t+C$

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

$v(0) = \frac{1}{3}(0)^3+\frac{3}{2}(0)^2+12(0)+C = 2$

Therefore, .

We can now use this in the velocity equation to give

$v(t) = \frac{1}{3}t^3+\frac{3}{2}t^2+12t+2$

integrating the velocity equation from to will give the distance

$\int_{5}^{7}\left[\frac{1}{3}t^3+\frac{3}{2}t^2+12t+2\right]dt \rightarrow$

$\rightarrow \frac{(2)^4}{12}+\frac{(2)^3}{2}+6(2)^2+2(2)+C- \frac{(1)^4}{12}-\frac{(1)^3}{2}-6(1)^2-2(1)-C$

$\rightarrow \frac{16}{12}+\frac{8}{2}+24+4- \frac{1}{12}-\frac{1}{2}-6-2 = 24.75$

Compare your answer with the correct one above

Question

Let be any point in the plane. What is the distanc between and its symmetric with respect to the origin?

Answer

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:

$d(a,b)=\sqrt{(x-(-x))^2+(y-(-y))^2}=\sqrt{4x^2+4y^2}$

and this gives: $d(a,b)=\sqrt{4}\sqrt{x^2+y^2}=2\sqrt{x^2+y^2}$ .

This shows the required result.

Compare your answer with the correct one above

Question

The velocity of a particle at time is given by the equation . How far does this particle travel from to ?

Answer

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

$x(t) = \int_{2}^{6}10t-7dt = 5(6)^2 - 7(6)-5(2)^2+7(2)$

Compare your answer with the correct one above

Question

Find the distance from the first point to the second point.

First point:

Second point:

Answer

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

$distance=\sqrt{x^2+y^2}$

So plugging in

$distance=\sqrt{2^2+2^2}$

$distance=\sqrt{4+4}$

$distance=\sqrt{8}$

$distance=2\sqrt{2}$

Compare your answer with the correct one above

Question

A race car is traveling at a constant 50 m/s when the driver suddenly hits the brakes. Assuming a constant deceleration of 10 m/s2, how far will the car travel before it comes to a complete stop?

Answer

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.

$a(t) = -10 m/s^{2}$

(the negative is used to represent deceleration).

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

$v(t) = ( -10 m/s^2 ) \cdot t + v_{0}$

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

$(-10 m/s^2 )\cdot t_{r} + 50 m/s = 0$

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.

$x(t_{r}) = x_{0} + v_{0}t_{r} + (1/2)at_{r}^2$

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:

$x(t_{r}) - x_{0} = v_{0}t_{r} + (1/2)at_{r}^2$

Plugging in our values for

$v_{0}:(50 m/s)$

and

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

$(x(t_{r}) - x_0)$ quantity, to be .

Compare your answer with the correct one above

Question

Assume that two students are each walking on a circular path. We further suppose that these paths are centered at the origin. The radius of the first one is , and the radius of the second is . One student stopped walking and is located at . What is the distance between the two students if other one is walking on the upper half of the circle of radius ?

Answer

We will use the distance formula to show this result. Recall that having two points with coordinates $(x_{1},y_{1}),(x_{2},y_{2})$ , then the distance between this two points is given by:

$\sqrt {(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$ . 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:

$y=\sqrt{1-x^2}$ ( We need only the upper half in this case , that is why y is $\ge 0$ .

Hence our second point is $(x,y)=(x,\sqrt{1-x^2})$ .

Using the distance formula we have: $\sqrt{(x-0)^2+(\sqrt{1-x^2}-2)^2)}$ .

We need to simplify this expression a bit:

$=\sqrt{x^2+4 -4\sqrt{1-x^2}+1-x^2}$

this gives finally after cancelling and adding:

$\sqrt{5-4\sqrt{1-x^2}$

Compare your answer with the correct one above

Question

Assume that two students are each walking on a circular path. We further suppose that these paths are centered at the origin. The radius of the first one is , and the radius of the second is . One student stopped walking and is located at . What is the distance between the two students if other one is walking on the lower half of the circle of radius ?

Answer

We will use the distance formula to show this result. Recall that having two points with coordinates $(x_{1},y_{1}),(x_{2},y_{2})$ , then the distance between this two points is given by:

$\sqrt {(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$ . 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:

$y=-\sqrt{1-x^2}$ ( We need only the lower half in this case , that is why y is $\le 0$ .

Hence our second point is $(x,y)=(x,-\sqrt{1-x^2})$ .

Using the distance formula we have: $\sqrt{(x-0)^2+(\sqrt{1-x^2}+2)^2)}$ .

We need to simplify this expression a bit:

$=\sqrt{x^2+4 +4\sqrt{1-x^2}+1-x^2}$

this gives finally after cancelling and adding:

$\sqrt{+4\sqrt{1-x^2}$

Compare your answer with the correct one above

Question

A person is sitting on and the other is walking on the line . What is the distance between the two?

Answer

We will use the distance formula to show this result. Recall that having two points with coordinates $(x_{1},y_{1}),(x_{2},y_{2})$ , then the distance between these two points is given by:

$\sqrt {(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$ .

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: $\sqrt{(x-1)^2+(4-1)^2}$ .

We need to simplify this expression a bit:

$\sqrt{(x^2-2x+1+9}$

this gives finally after cancelling and adding:

$\sqrt{x^2-2x+10}$

Compare your answer with the correct one above

Question

Joe is sitting on , and James is walking along the line .

What is the distance between the two ?( As a function of )

Answer

We will use the distance formula to show this result. Recall that having two points with coordinates $(x_{1},y_{1}),(x_{2},y_{2})$ , then the distance between these two points is given by:

$\sqrt {(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$ .

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: $\sqrt{(x-0)^2+(7-0)^2}$ .

We need to simplify this expression a bit to get:

$\sqrt{x^2+49}$

Compare your answer with the correct one above

Question

A person is restricted to sit on the center of the circle:

.

Another person is moving along the line . What is the distance between the two at any given position of the second?

Answer

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 :

$\sqrt{(x-1)^2 +(4-1)^2}$ .

We can simplify this expression to get :

$\sqrt{x^2-2x+1+9}=\sqrt{x^2-2x+10}$

Compare your answer with the correct one above

Question

The line intersects the unit circle at two points. What is the distance between these points?

Answer

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:

$\left{\begin{matrix} x^2+y^2=1\ y+2x-1=0\end{matrix}\right.$ which we can write by solving for y in the second equation.

$\left{\begin{matrix} x^2+y^2=1\ y=1-2x \end{matrix}\right.$

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

. This gives:

and hence we have:

.

This gives or $x=\frac{ 4}{5}$ .

We deduce the value of y from:

.

If .

If $x=\frac{4}{5}, y=-\frac{3}{5}$

We have the two points of interesection: $(0,1);\left(\frac{4}{5},\frac{-3}{5}\right)$ . Now we use the distance formula to the distance :

$\sqrt{\left(\frac{4}{5}\right)^2+\left(\frac{8}{5}\right)^2}=\frac{\sqrt{80}}{5}=\frac{4\sqrt5}{5}$

Compare your answer with the correct one above

Tap the card to reveal the answer