### All Calculus 1 Resources

## Example Questions

### Example Question #1 : Distance

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

**Possible Answers:**

**Correct answer:**(1,2,3)

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)

### Example Question #2 : Distance

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?

**Possible Answers:**

647.12 m

None of the other answers

64.65 m

20.42 m

44.24 m

**Correct answer:**

44.24 m

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 + c_{1}

To solve for c_{1}, we know that at t = 0, the initial velocity was 4. This means v(0) = sin(0) + 6 * 0 + c_{1} = 4; sin(0) = 0, so we know c_{1} = 4

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

s(t) = –cos(t) + 3t^{2} + 4 * t + c_{2 }(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*4^{2} + 16 + c_{2} = –cos(4) + 48 + 16 + c_{2 }= –cos(4) + 64 + c_{2}

s(2) = –cos(2) + 3*2^{2} + 8 + c_{2 }= –cos(2) + 12 + 8 + c_{2} = –cos(2) + 20 + c_{2}

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

### Example Question #3 : Distance

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?

**Possible Answers:**

84

None of the other answers

104

20

64

**Correct answer:**

84

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

∫[4, 8](4t – 3)dt = [4, 8](2t^{2} – 3t) = (2 * 8^{2} – 3 * 8) – (2 * 4^{2} – 3 * 4)

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

### Example Question #4 : Distance

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.

**Possible Answers:**

**Correct 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:

### Example Question #741 : Calculus

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:

**Possible Answers:**

**Correct 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:

### Example Question #1 : How To Find Distance

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

**Possible Answers:**

None of these answers are correct.

**Correct answer:**

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.

### Example Question #743 : Calculus

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 .

**Possible Answers:**

**Correct answer:**

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

### Example Question #6 : Distance

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

**Possible Answers:**

**Correct answer:**

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

### Example Question #7 : Distance

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

First point:

Second point:

**Possible Answers:**

**Correct 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

So plugging in

### Example Question #746 : Calculus

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 ?

**Possible Answers:**

**Correct answer:**

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:

### All Calculus 1 Resources

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