When using calculus for useful applications, the equations and subsequent derivatives usually mean something or describe something. On this page, we discuss the situation when a function represents the position of an object, in two dimension motion, vertically, horizontally or a combination. Beginning calculus topics are the only requirements to understand this material.
If you are looking for more advanced discussions, see the linear motion pages involving integration or differential equations.
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Usually we derive or start with a function whose variable is time. This usually looks like \( s(t) \) where the independent variable is t and represents time. The function itself is called a position function. This means that we can plug in a time, like \( t=3 \), to the function, \( s(3) \) and the result is the (instantaneous) position of the object at time \( t=3 \). Of course the units are dependent upon what the equation represents. Also, we usually assume that time starts at \(t=0\) and always increases.
When we have a position function, the first two derivatives have specific meanings. The first derivative is the velocity and the second derivative is the acceleration of the object. We take the derivative with respect to the independent variable, t.
The units of velocity are distance per unit time, in MKS units, meters per second, m/s.
The units of acceleration are distance per unit time squared, in MKS units, meters per second squared, m/s^{2}.
Note About Velocity - - There is a difference between velocity and speed. Velocity can be negative and includes direction information. Speed is the magnitude of velocity and is always positive. Speed does not include direction. If you are familiar with vectors, velocity is a vector, speed is a scalar whose value is the magnitude of a velocity vector. Many people use these terms interchangable, which is incorrect.
Here are the equations you need.
Equations | ||
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position | \(s(t) = at^2/2 + v_0t + s_0\) | distance |
velocity | \(s'(t) = v(t) = at + v_0\) | distance per unit time |
acceleration | \(s''(t) = a(t) = a\) | distance per unit time squared |
Note About Acceleration - - In the above table, we assumed acceleration is constant. This is true in many problems, especially ones where we talk about the acceleration due to gravity near the surface of the earth, for example. However, this is not true in ALL problems. So you need to pay attention to what is going on in the problem statement if you are required to set up these equations.
Constants and Variables | |
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\(s_0\) | initial position; can also be written \(s(0)\) |
\(v_0\) | initial velocity; can also be written \(v(0)\) |
\(a\) | acceleration is often, but not always, constant |
\(t\) | independent variable time |
Okay, let's watch a video to get this into our heads a little bit more. Here is a good video that starts out by giving an overview of the derivative relationship between these equations. Then he does a quick example, and then finishes by giving a more intuitive description of the relationships. He does talk a little bit about anti-derivatives but if you aren't there yet, this video will still help you.
video by PatrickJMT |
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Your best course of action now is to work as many practice problems as possible.
Practice
Basic |
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\(s(t)=3t^2+5t+2\) describes a particle's motion with units in meters. Find (a) the velocity and acceleration functions, and
(b) the instantaneous velocity and acceleration at \(t=4\) sec.
Problem Statement |
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\(s(t)=3t^2+5t+2\) describes a particle's motion with units in meters. Find (a) the velocity and acceleration functions, and
(b) the instantaneous velocity and acceleration at \(t=4\) sec.
Final Answer |
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a) \(v(t) = 6t+5~ m/s\) \(a(t) = 6~ m/s^2\)
b) \(v(4)=29~ m/s\) \(a(4) = 6~ m/s^2\)
Problem Statement |
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\(s(t)=3t^2+5t+2\) describes a particle's motion with units in meters. Find (a) the velocity and acceleration functions, and
(b) the instantaneous velocity and acceleration at \(t=4\) sec.
Solution |
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For this problem, we can tell that the position is in meters from the definition of \(s(t)\) and we infer that time is in seconds from part b of the problem statement.
a) Since the velocity is the first derivative and the acceleration is the second derivative, we have the following.
\(v(t)=s'(t) = 6t+5 \)
\(a(t)=v'(t)=s''(t) = 6 \)
b) The instantaneous velocity and acceleration are given by letting \(t=4s\) in the equations we found in part a.
\(v(4) = 6(4)+5 = 24+5=29\)
\(a(4) = 6\)
Final Answer |
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a) \(v(t) = 6t+5~ m/s\) \(a(t) = 6~ m/s^2\)
b) \(v(4)=29~ m/s\) \(a(4) = 6~ m/s^2\)
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A particle moves along a path with its position \(p=t^2-6t+8\) meters. a) When does the velocity equal 30 meters per second? b) When is the particle at rest?
Problem Statement |
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A particle moves along a path with its position \(p=t^2-6t+8\) meters. a) When does the velocity equal 30 meters per second? b) When is the particle at rest?
Solution |
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He works this problem using the limit definition of the derivative, which is good practice but you can also work it by just taking the derivative. Of course, check with your instructor to see what they require.
video by PatrickJMT |
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If the position of a particle is given by \(x(t)=100-16t^2\), find the position of the particle when the velocity is zero.
Problem Statement |
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If the position of a particle is given by \(x(t)=100-16t^2\), find the position of the particle when the velocity is zero.
Final Answer |
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\(x = 100\)
Problem Statement |
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If the position of a particle is given by \(x(t)=100-16t^2\), find the position of the particle when the velocity is zero.
Solution |
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video by Krista King Math |
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Final Answer |
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\(x = 100\)
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A car with position function \(x(t)=100t-5t^2\) is traveling \(100~ft/s\) when the driver suddenly applies the brakes. How far and for how long does the car skid before it comes to a stop?
Problem Statement |
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A car with position function \(x(t)=100t-5t^2\) is traveling \(100~ft/s\) when the driver suddenly applies the brakes. How far and for how long does the car skid before it comes to a stop?
Final Answer |
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skids 500ft for 10secs
Problem Statement |
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A car with position function \(x(t)=100t-5t^2\) is traveling \(100~ft/s\) when the driver suddenly applies the brakes. How far and for how long does the car skid before it comes to a stop?
Solution |
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video by Krista King Math |
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Final Answer |
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skids 500ft for 10secs
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The vertical position of a ball is given by \(y(t)=-16t^2+96t+50 ft\). What is the maximum height the ball will reach?
Problem Statement |
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The vertical position of a ball is given by \(y(t)=-16t^2+96t+50 ft\). What is the maximum height the ball will reach?
Final Answer |
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max height 194ft
Problem Statement |
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The vertical position of a ball is given by \(y(t)=-16t^2+96t+50 ft\). What is the maximum height the ball will reach?
Solution |
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video by Krista King Math |
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Final Answer |
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max height 194ft
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Intermediate |
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A ball with position function \(y(t)=-gt^2/2+v_0t+s_0\) is thrown straight upward from the ground with an initial velocity of 96 ft/sec. Find the maximum height that the ball attains and its velocity when it hits the ground.
Problem Statement |
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A ball with position function \(y(t)=-gt^2/2+v_0t+s_0\) is thrown straight upward from the ground with an initial velocity of 96 ft/sec. Find the maximum height that the ball attains and its velocity when it hits the ground.
Final Answer |
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max height \(144~ft\)
velocity when it hits the ground \(-96~ft/sec\)
Problem Statement |
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A ball with position function \(y(t)=-gt^2/2+v_0t+s_0\) is thrown straight upward from the ground with an initial velocity of 96 ft/sec. Find the maximum height that the ball attains and its velocity when it hits the ground.
Solution |
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video by Krista King Math |
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Final Answer |
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max height \(144~ft\)
velocity when it hits the ground \(-96~ft/sec\)
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The motion of a ball is described by \(y(t)=-16t^2+64t\). Find the maximum height and the velocity when the ball hits the ground.
Problem Statement |
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The motion of a ball is described by \(y(t)=-16t^2+64t\). Find the maximum height and the velocity when the ball hits the ground.
Final Answer |
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max height \(64~ft\)
velocity \(-64~ft/sec\)
Problem Statement |
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The motion of a ball is described by \(y(t)=-16t^2+64t\). Find the maximum height and the velocity when the ball hits the ground.
Solution |
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video by Krista King Math |
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Final Answer |
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max height \(64~ft\)
velocity \(-64~ft/sec\)
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A coin is dropped from the roof of a 600ft tall building with initial velocity of \(-8 ft/sec\). When does it hit the ground and what is its velocity at that point?
Problem Statement |
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A coin is dropped from the roof of a 600ft tall building with initial velocity of \(-8 ft/sec\). When does it hit the ground and what is its velocity at that point?
Solution |
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video by Krista King Math |
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Really UNDERSTAND Calculus
external links you may find helpful |
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The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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