All Calculus 1 Resources
Example Questions
Example Question #331 : Acceleration
The position of a is given by the following functions:
Find the acceleration.
Answer not listed
In order to find the acceleration of a certain point, you first find the derivative of the position function to get the velocity function and the derivative of the velocity function to get the acceleration function:
In this case, the position function is:
Then take the derivative of the position function to get the velocity function:
Then take the derivative of the velocity function to get the acceleration function:
Then, plug into the acceleration function:
Therefore, the answer is:
Example Question #341 : How To Find Acceleration
The position of a is given by the following functions:
Find the acceleration.
Answer not listed
In order to find the acceleration of a certain point, you first find the derivative of the position function to get the velocity function and the derivative of the velocity function to get the acceleration function:
In this case, the position function is:
Then take the derivative of the position function to get the velocity function:
Then take the derivative of the velocity function to get the acceleration function:
Then, plug into the acceleration function:
Therefore, the answer is:
Example Question #341 : How To Find Acceleration
Given the position function of an object in motion (in meters), find the acceleration of the object at t=2 seconds.
-4.9 m/s/s
4.9 m/s/s
None of the other answers.
-9.8 m/s/s
9.8 m/s/s
-9.8 m/s/s
The acceleration of an object in motion at any given time is modeled by the second derivative of its position function. That is
The acceleration of the object at t=2 seconds is then
In this case, the acceleration is constant across all values of t.
Example Question #342 : How To Find Acceleration
Find the acceleration of a particle at given the following velocity function:
The first step is to obtain the acceleration equation.
This is done by taking the first derivative of the velocity function,
using the power rule,
.
The acceleration equation
can be used to find the acceleration at any given time.
In our case, plugging in 5 for t into the equation gives,
.
Example Question #341 : How To Find Acceleration
Find the acceleration of a particle given its velocity function is .
The acceleration function can be found given any velocity function by taking its derivative.
Using the power rule
,
the derivative becomes,
.
Recall that the derivative of a constant is always zero.
Example Question #342 : How To Find Acceleration
Find the acceleration function of a car who's velocity function is given by .
The derivative of the velocity function is acceleration, and using the power rule
,
you get
.
Recall that the derivative of a constant is zero.
Example Question #1 : How To Find Distance
Find the midpoint of the line segment connecting the points (-1,0,5) and (3,4,1).
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 #1 : 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?
None of the other answers
44.24 m
20.42 m
64.65 m
647.12 m
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 + c1
To solve for c1, we know that at t = 0, the initial velocity was 4. This means v(0) = sin(0) + 6 * 0 + c1 = 4; sin(0) = 0, so we know c1 = 4
Therefore, v(t) = sin(t) + 6t + 4
s(t) = –cos(t) + 3t2 + 4 * t + c2 (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*42 + 16 + c2 = –cos(4) + 48 + 16 + c2 = –cos(4) + 64 + c2
s(2) = –cos(2) + 3*22 + 8 + c2 = –cos(2) + 12 + 8 + c2 = –cos(2) + 20 + c2
s(4) - s(2) = –cos(4) + 64 + cos(2) – 20 = 44.237496784317 (approx.) or 44.24
Example Question #2 : 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?
64
20
None of the other answers
104
84
84
To find the change in position, we must take the definite integral of the velocity function
∫[4, 8](4t – 3)dt = [4, 8](2t2 – 3t) = (2 * 82 – 3 * 8) – (2 * 42 – 3 * 4)
= 2 * 64 – 24 – (2 * 16 – 12) = 104 – 20 = 84
Example Question #731 : Spatial Calculus
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.
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: