### All Calculus 1 Resources

## Example Questions

### Example Question #1 : Solving For Time

The initial position of a particle is 0. If its velocity is described by v(t) = 4t + 5, what is its position at the time when the velocity is equal to 45?

**Possible Answers:**

250

105

125

90

173

**Correct answer:**

250

The position s(t) is equal to ∫v(t) dt = ∫ 4t + 5 dt = 2t^{2} + 5t + C

Now, since we know that the initial position of the particle (at t = 0) is 0, we know C is 0. Therefore, s(t) = 2t^{2} + 5t + C

To find the time t when at which the velocity is 45, set v(t) equal to 45. 45 = 4t + 5 → 40 = 4t → t = 10

The position of the particle is s(10) = 2 * 10^{2} + 50 = 200 + 50 = 250

### Example Question #1831 : Functions

The position of a particle is defined by s(t) = 4t^{3} – 3t^{2} + 2t. At what time (to the nearest hundreth) is its velocity equal to 384?

**Possible Answers:**

15 ± √(3)

(3 ± √(4593))/12

(22 ± √(5))/10

48 ± √(17)

(4 ± √(4721))/2

**Correct answer:**

(3 ± √(4593))/12

The position of a particle is defined by s(t) = 4t^{3} – 3t^{2} + 2t. At what time (to the nearest hundreth) is its velocity equal to 384?

Start by finding the velocity function:

v(t) = s'(t) = 12t^{2} – 6t + 2

To find the time, set v(t) equal to 384: 384 = 12t^{2} – 6t + 2

To solve, set equal the equation equal to 0: 12t^{2} – 6t – 382 = 0

Use the quadratic formula:

(–(–6) ± √(36 – 4*12*(–382)))/(2 * 12)

= (6 ± √(36 + 18336))/24 = (6 ± √(18372))/24 = (6 ± 2√(4593))/24

= (3 ± √(4593))/12

### Example Question #3 : Solving For Time

The initial position of a particle is 44.5. If its velocity is described by v(t) = 3t + 12, what is its position at the time when the velocity is equal to 8391?

**Possible Answers:**

11734745

11524231

11734834

1038181

11734789.5

**Correct answer:**

11734834

The initial position of a particle is 44.5. If its velocity is described by v(t) = 3t + 12, what is its position (to the nearest hundreth) at the time when the velocity is equal to 8391?

To solve for t, set v(t) equal to 8391: 3t + 12 = 8391; 3t = 8379; t = 2793

Now, the position function is equal to ∫v(t)dt = ∫ 3t + 12 dt = (3/2)t^{2} + 12t + C

Now, since the initial position is 44.5, C is 44.5; therefore, s(t) = (3/2)t^{2} + 12t + 44.5

To solve for the position, solve s(2793) = (3/2)2793^{2} + 12*2793 + 44.5 = 1.5 * 7800849 + 33516 + 44.5 = 11701273.5 + 33516 + 44.5

### Example Question #4 : Solving For Time

An eagle flies at a parabolic trajectory such that , where is in the height in meters and is the time in seconds. At what time will its velocity ?

**Possible Answers:**

**Correct answer:**

Take the derivative of the position function to obtain the velocity function.

We want to know the time when the velocity is -8. Substitute v into the equation to find t.

### Example Question #1 : How To Solve For Time

A ball is thrown in the air, modeled by the function . At what time will the ball hit the ground?

**Possible Answers:**

**Correct answer:**

To find the time when the ball hits the ground, set and solve for .

Separate each term and solve for t.

Since negative time does not exist, the only possible answer is .

### Example Question #6 : Solving For Time

Suppose the acceleration function of a biker going uphill at the start of a race is , where is in seconds. When will it take the biker to reach constant velocity?

**Possible Answers:**

**Correct answer:**

Constant velocity means there is neither an increase or decrease in acceleration.

Substitute acceleration and solve for time.

### Example Question #7 : Solving For Time

For this problem, the acceleration of gravity is simplified to .

I fire a cannon straight into the air. Assuming that the cannonball leaves the cannon at a velocity of , weighs kgs, and is fired from the ground (i.e. m), how long will it take before the cannonball reaches the ground again?

**Possible Answers:**

**Correct answer:**

Initial velocity is given as 100 m/s and the acceleration due to gravity is towards the Earth , or .

We want to find how long it will take for the velocity of the cannonball to reach , so we set , where t is time and .

So, , , .

Therefore, it takes ten seconds for the ball to reach a velocity of 0, and given that acceleration is uniform (i.e. a constant), we know that it will take the same amount of time to come down as it took to go up, or ten seconds. Therefore, the total time the cannonball spends in the air is,

seconds.

### Example Question #8 : Solving For Time

Given the instantaneous velocity and the position function, find the time at which the moving object reaches that instantaneous velocity.

**Possible Answers:**

**Correct answer:**

We begin by finding the derivative of the position function using the power rule:

We then set the given instantaneous velocity equal to the velocity function and solve for t:

### Example Question #9 : Solving For Time

Assume the acceleration due to gravity is

If you throw a ball straight up with an initial velocity of how many seconds will it take before the ball returns to your hand?

**Possible Answers:**

**Correct answer:**

Write out the equation for the height of the ball

.

You can arrive at this by starting with the information that the acceleration on the ball is the constant acceleration due to gravity and integrating twice. Making sure to solve for your constants along the way.

Your initial velocity of and your initial position of will help write out this equation.

Then solve for the two values of for which the ball is at height . Those are seconds and seconds.

### Example Question #10 : Solving For Time

A ball is thrown upwards at a speed of from a building. Assume gravity is .

Which of the following is closest to the time after the initial throw before the ball hits the ground?

**Possible Answers:**

**Correct answer:**

If we approximate gravity as we can simplify into and use the quadratic formula to find the time at which the position of the ball is zero (the ball hits the ground).

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