All Calculus 1 Resources
Example Questions
Example Question #261 : Calculus
Suppose the acceleration of a particle is two at .  The acceleration of the particle is three whenÂ
.  What is the velocity atÂ
 if the velocity atÂ
 is zero?
By knowing the two accelerations at two different times, it is possible to find the acceleration function.  Use the slope and slope-intercept equation to determine the acceleration function.  The two points, where acceleration is the dependent variable , areÂ
 andÂ
.
Find the slope.
Find the y-intercept of the acceleration function and choose a point for substitution.
The acceleration function is:
Integrate this function to obtain the velocity.
Substitute the initial condition where velocity is zero when the time is at zero.
Write the velocity function.
Substitute  into the velocity function.
Example Question #261 : Spatial Calculus
Find velocity at  given the position function below.
To solve, simply differentiate the position function to find the velocity function and then plug in . Thus,
Example Question #263 : Calculus
Find the velocity at  given the position equation below.
To solve, simply differentiate and then plug in  for
.
Example Question #262 : Spatial Calculus
Find the velocity at  given the position function below.
To solve, simply differentiate once and then plug in . Thus,
Example Question #263 : Spatial Calculus
Find the velocity of the particle if the acceleration of the particle is given by the following function:
To find the velocity function from the acceleration function, we must integrate:
The integral was performed using the following rule:
Example Question #264 : Spatial Calculus
Let  describe the position of a particle traveling along the
 axis where each unit is
 inch.
Find the velocity of the particle at  seconds.
There is not enough information given to determine the answer.
Velocity is the derivative of position, so we need to take the derivative of our equation before plugging in .
For this particular function we will need to apply the power rule to find the derivative.
The power rule states, .
Also recall that the derivative of a constant is always zero.
Applying the power rule once to our position function we will find the following velocity function.
.
Thus, .
From here, substitute  to solve for the velocity in question.
Â
Example Question #267 : Calculus
A car is traveling down a straight road, and its position (in miles from the starting point) can be described as a function of time (in minutes traveled) by the equationÂ
.
When is the car traveling at ?
After  minutes
After  minutes
There is not enough information given to determine the answer.
After  minutes
After  minutes
After  minutes
To solve this problem we need identify what the question is asking for. This particular question is asking us to find the time for which the car is traveling a certain speed. Recall that speed is another term for velocity. Therefore, we will need to find the velocity function, set it equal to the specific speed, and solve for time. Since we are given the position function, we need to take the derivative once to find the velocity function.
For this particular function we will need to use the power rule which states, .
Also recall the derivative of a constant is always zero.
IfÂ
,
then by applying the power rule we get,
.
Setting our velocity function equal to the specific velocity given in the equation we now have,
and finally  minutes.
Example Question #268 : Calculus
An arrow that is shot straight up in the air can be located using the general position functionÂ
.
The initial velocity of the arrow is .
Let the acceleration due to gravity be .
What is the velocity of the arrow when it returns to the ground?
There is not enough information to determine the solution.
We can set the position function equal to zero to determine when the arrow is on the ground.Â
The arrow is at ground level at  andÂ
.
From here, we want to find the velocity function by taking the derivative of the position function. For this particular function we will need to use the power rule to find the velocity function.
The power rule states, .
Applying the above rule we find the velocity to be,
.
Substituting  we are able to find the velocity of the arrow when it hits the ground.
Example Question #269 : Calculus
While struggling with your spaghetti at dinner one evening, you accidentally shove a meatball off your plate; it rolls a full  across the table before stopping. If the meatball decelerated at a constant rate of
, what was its initial velocity?
There is not enough information given to determine the answer.
First, we can list the things we know from the problem.Â
, we use a negative because the meatball is slowing down.
Let  be the distance the meatball has traveled. Then we know
. We do not know how many seconds the meatball rolled, so we can call the final time the meatball stops
. Now
.
The meatball stops rolling at time  as well, so
. We need to determine
. We do not know enough about position yet to determine velocity, so let's work in reverse. We begin withÂ
 and integrate to determine velocity.
When integrating recall the rule,
.
We do not have enough information about the velocity to determine , so let's keep going and look at position.
We have , so we can use this to determine the constant
.
.Â
We now have a system of equations for position and velocity that both share the constant .
We have information about what happens at time  for both of these, so we can use that here.
We can use substitution to solve this system.
Because  is a particular time, a negative value does not make sense. Hence, the meatball stops rolling after
.
Now, we simply need to find the value of  and the initial velocity.
Example Question #270 : Calculus
The position of a particle traveling along the  axis is described byÂ
.
When is the particle at rest?
The particle is never at rest.
At approximatelyÂ
At approximatelyÂ
At approximatelyÂ
There is not enough information given to determine the answer.
At approximatelyÂ
When the particle is at rest, its velocity will be . We can use the position function to find the velocity.
Velocity is the derivative of position. In this particular function we will use the power rule to differentiate.
The power rule states, .
Also recall that the derivative of a constant is always zero.
Applying the above rules we find the following velocity function.
Now set the velocity function equal to zero and solve for time.
Using the quadratic formula,
whereÂ
we findÂ
.
Because a negative value for time does not make sense, we know  seconds.
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