### All Calculus 1 Resources

## Example Questions

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

If during the first 15 seconds of its flight the displacement of a spacecraft is given by the equation 2^{1.25x} + x^{2 }feet, what is its velocity at 10 seconds, given that its initial velocity is 0?

**Possible Answers:**

1.25 * 2^{12.5} * log(2) + 20 feet/second

None of the other answers

1.25 * 2^{12.5} * log(2) + 100 feet/second

2^{1.25} + 20 feet/second

1.25 * 2^{12.5} + 20 feet/second

**Correct answer:**

1.25 * 2^{12.5} * log(2) + 20 feet/second

To find the velocity function based on displacement, use the first derivative of f(x) = 2^{1.25x} + x^{2}. f'(x) = 1.25 * ln(2) * 2^{1.25x} + 2x

The velocity at x = 10 would therefore be: f'(10) = 1.25 * log(2) * 2^{1.25 * 10} + 2 * 10 = 1.25 * 2^{12.5} * log(2) + 20 feet/second

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

What is the instantaneous velocity at time t = π/2 of a particle whose positional equation is represented by s(t) = 12tan(t/2 + π)?

**Possible Answers:**

6√2

12

3√2

12√2

6

**Correct answer:**

12

The instantaneous velocity is represented by the first derivative of the positional equation.

v(t) = s'(t) = 12 * (1/2) sec^{2}(t/2 + π) = 6sec^{2}(t/2 + π) = 6/cos^{2}(t/2 + π) = 6/((–1)^{2} cos^{2}(t/2)) = 6/cos^{2}(t/2)

Based on the nature of the cosine, we know that 6/cos^{2}(t/2 + π) = 6/((–1)^{2} cos^{2}(t/2)) = 6/cos^{2}(t/2)

To solve, v(π/2) = 6/cos^{2}(t/2) = 6/((1/√2)^{2}) = 6/(1/2) = 12

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

What is the instantaneous velocity at time t=π/2 of a particle whose positional equation is represented by s(t) = 12cos^{2}(t/2 + π)?

**Possible Answers:**

0

2√3

–12

–6

2

**Correct answer:**

–6

The instantaneous velocity is represented by the first derivative of the positional equation. This is found by using the chain rule both on the square of the cosine function and the function itself.

v(t) = s'(t) = 12 * 2 cos(t/2 + π) * (–sin(t/2 + π)) * (1/2) = –12cos(t/2 + π)sin(t/2 + π)

Given what we know about the cosine and sine functions, we know cos(t/2 + π) = –cos(t/2) and sin(t/2 + π) = –sin(t/2)

Therefore, v(t) = –12(–sin(t/2))(–cos(t/2)) = –12sin(t/2)cos(t/2)

v(π/2) = –12sin(π/4)cos(π/4) = –12(1/√2)(1/√2) = –12 * (1/2) = –6

### Example Question #1 : Calculus

A weight hanging from a spring is stretched down 3 units beyond its rest position and released at time *t*=0 to bob up and down. Its position at any later time *t* is

What is its velocity at time ?

**Possible Answers:**

**Correct answer:**

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

The position of a particle at time is given by . What is the particle's velocity at time

**Possible Answers:**

**Correct answer:**

The velocity function is given by the derivative of the position function. So here . Plugging 3 in for gives 16.

### Example Question #3 : Calculus

The position of a particle is given by . Find the velocity at .

**Possible Answers:**

**Correct answer:**

The velocity is given as the derivative of the position function, or

.

We can use the quotient rule to find the derivative of the position function and then evaluate that at . The quotient rule states that

.

In this case, and .

We can now substitute these values in to get

.

Evalusting this at gives us .

So the answer is .

### Example Question #4 : Calculus

Find the velocity function if the position function is given as: .

**Possible Answers:**

**Correct answer:**

There are three terms in this problem that has to be derived. The derivative of the position function, or the velocity function, represents the slope of the position function.

The derivative of can be solved by using the power rule, which is:

Therefore. the derivative of is .

The derivative of is by using the constant multiple rule.

The derivative of is since derivatives of constants equal to zero.

### Example Question #5 : Calculus

Find the velocity function given the position function: .

**Possible Answers:**

**Correct answer:**

The derivatives can be solved term by term.

First, find the derivative of . This can be done by the power rule.

Find the derivative of . Rewrite this as to compute by power rule.

Therefore, the velocity function is:

### Example Question #6 : Calculus

Find the velocity of a function if the acceleration is: .

**Possible Answers:**

**Correct answer:**

To find the velocity given the acceleration function, we will need to integrate the acceleration function.

### Example Question #7 : Calculus

Find the velocity at if the acceleration function is: .

**Possible Answers:**

**Correct answer:**

The velocity function can be obtained by integrating the acceleration function.

Since we are finding the velocity at , substitute this into the velocity function.

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