Differentiation

NZ Level 7 (NZC) Level 2 (NCEA)

Speed, displacement and velocity

Lesson

Differentiation holds the key to a beautiful mathematical relationship between Speed and Distance, and Velocity and Displacement.

Distance - a scalar quantity that describes how far an object is from a fixed point. As this is a distance it is measured in units of length such as km, m, cm.

Speed - a scalar quantity that describes the rate of change of an object, measured using a distance/time unit such as km/h, or m/s. We usually use $s(t)$`s`(`t`) to describe a speed function describing speed over time.

Displacement - a vector quantity that describes the distance an object is from a fixed point with its corresponding direction. We usually use $x(t)$`x`(`t`) to describe a displacement function describing displacement over time.

Velocity - a vector quantity that describes the speed of an object and it's corresponding direction. We usually use $v(t)$`v`(`t`) to describe a velocity function describing velocity over time.

The derivative of a distance function gives you a function to determine the speed.

So if the distance of an object can be described using $d(t)=t^2+4t$`d`(`t`)=`t`2+4`t`, then the speed of the object can be described using $\frac{dd}{dt}=s(t)=2t+4$`d``d``d``t`=`s`(`t`)=2`t`+4.

The derivative of a displacement function gives you a function to determine the velocity.

So if the displacement of an object can be described using $x(t)=t^3-2t^2+5$`x`(`t`)=`t`3−2`t`2+5, then the velocity of the object can be described using $\frac{dx}{dt}=v(t)=3t^2-4t$`d``x``d``t`=`v`(`t`)=3`t`2−4`t`.

You may have noticed how similar the above two statements are. In fact the theory is identical, the only difference is that when working with distance we get speed, and when working with displacement we get velocity. Also, there is such a thing as the second derivative (and third, and fourth etc), and we will find out more about all of these later but interestingly if we take the derivative of $v(t)$`v`(`t`), which is the derivative of displacement, we get an acceleration function. We will also look at these later, but thought you'd like to know where we might be heading with all of this!

The position (in metres) of an object along a straight line after $t$`t` seconds is modelled by $x\left(t\right)=6t^2$`x`(`t`)=6`t`2.

State the velocity $v\left(t\right)$

`v`(`t`) of the particle at time $t$`t`.Which of the following represent the velocity of the particle after $4$4 seconds? Select all that apply.

$x'\left(4\right)$

`x`′(4)A$v'\left(4\right)$

`v`′(4)B$x\left(4\right)$

`x`(4)C$v\left(4\right)$

`v`(4)D$x'\left(4\right)$

`x`′(4)A$v'\left(4\right)$

`v`′(4)B$x\left(4\right)$

`x`(4)C$v\left(4\right)$

`v`(4)DHence find the velocity of the particle after $4$4 seconds.

The position (in metres) of an object along a straight line after $t$`t` seconds is modelled by $x\left(t\right)=18\sqrt{t}$`x`(`t`)=18√`t`.

Determine the function $v\left(t\right)$

`v`(`t`) for the velocity of the particle. Express $v\left(t\right)$`v`(`t`) in surd form.Hence, calculate the velocity of the object after $9$9 seconds.

The displacement of a particle moving in rectilinear motion is given by $x\left(t\right)=-5t\left(t-4\right)$`x`(`t`)=−5`t`(`t`−4) where $x$`x` is the displacement in metres from the origin and $t$`t` is the time in seconds.

Calculate the initial displacement of the particle.

Solve for the time $t$

`t`when the particle next returns to the origin.Using graphical methods, calculate the distance traveled by the particle between leaving the origin and returning again.

Apply differentiation and anti-differentiation techniques to polynomials

Apply calculus methods in solving problems