The position s(t) is equal to ∫v(t) dt = ∫ 4t + 5 dt = 2t² + 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) = 2t² + 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 * 10² + 50 = 200 + 50 = 250

The position of a particle is defined by s(t) = 4t³ – 3t² + 2t.  At what time (to the nearest hundreth) is its velocity equal to 384?

Start by finding the velocity function:

v(t) = s'(t) = 12t² – 6t + 2

To find the time, set v(t) equal to 384: 384 = 12t² – 6t + 2

To solve, set equal the equation equal to 0: 12t² – 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 

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)t² + 12t + C

Now, since the initial position is 44.5, C is 44.5; therefore, s(t) = (3/2)t² + 12t + 44.5

To solve for the position, solve s(2793) = (3/2)2793² + 12*2793 + 44.5 = 1.5 * 7800849 + 33516 + 44.5 = 11701273.5 + 33516 + 44.5

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.

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 .

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

Substitute acceleration and solve for time.

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.

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:

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.

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).