NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Velocity and acceleration as functions of x

Lesson

If $y$`y` is a smooth function of $t$`t` and $x$`x` is also a function of $t$`t`, we can obtain $\frac{\mathrm{d}y}{\mathrm{d}t}$`d``y``d``t` and also $\frac{\mathrm{d}x}{\mathrm{d}t}$`d``x``d``t` and then find $\frac{\mathrm{d}y}{\mathrm{d}x}$`d``y``d``x` from the relation

$\frac{\mathrm{d}y}{\mathrm{d}t}=\frac{\mathrm{d}y}{\mathrm{d}x}.\frac{\mathrm{d}x}{\mathrm{d}t}$`d``y``d``t`=`d``y``d``x`.`d``x``d``t`

In kinematics, we make use of the concepts *displacement*, *velocity** *and *acceleration *when describing the motion of a particle or body. Strictly speaking, these are all vector quantities since each is specified by a direction with respect to a coordinate system as well as a magnitude. For simplicity, we consider here just one available direction - along the $x$`x`-axis.

The three quantities are variable and are usually understood to be functions of time. Thus, we have a displacement function $x(t)$`x`(`t`), a velocity function $v(t)$`v`(`t`) and an acceleration function $a(t)$`a`(`t`).

These are related by

$v(t)=\frac{\mathrm{d}x}{\mathrm{d}t}$`v`(`t`)=`d``x``d``t` and

$a(t)=\frac{\mathrm{d}v}{\mathrm{d}t}$`a`(`t`)=`d``v``d``t`

It can be useful to write the acceleration function as a function of the displacement $x$`x` rather than as a function of time. This is done by means of the chain rule. Thus,

$a(t)=\frac{\mathrm{d}v}{\mathrm{d}t}=\frac{\mathrm{d}v}{\mathrm{d}x}.\frac{\mathrm{d}x}{\mathrm{d}t}$`a`(`t`)=`d``v``d``t`=`d``v``d``x`.`d``x``d``t`

But, since $\frac{\mathrm{d}x}{\mathrm{d}t}=v$`d``x``d``t`=`v`, we have

$a(t)=v\frac{\mathrm{d}v}{\mathrm{d}x}$`a`(`t`)=`v``d``v``d``x`

After $t$`t` seconds a body has displacement $x$`x` metres, velocity $v\ \text{m/s}$`v` m/s and acceleration $a\ \text{m/s}^2$`a` m/s2.

The velocity increases with displacement according to $v=\log_e\ x$`v`=`l``o``g``e` `x`. Express the acceleration as a function of $x$`x`.

Since $a(t)=\frac{\mathrm{d}v}{\mathrm{d}t}$`a`(`t`)=`d``v``d``t` we have $a(t)=v\frac{\mathrm{d}v}{\mathrm{d}x}=\frac{v}{x}$`a`(`t`)=`v``d``v``d``x`=`v``x`. We see that the acceleration is decreasing as displacement increases.

After $t$`t` seconds, the position of a particle from a fixed point is $x$`x` metres, its velocity is $v$`v` m/s and its acceleration is $a$`a` m/s^{2}.

If $v=x+3$`v`=`x`+3, express $a$`a` in terms of $x$`x`.

After $t$`t` seconds, the position of a particle is $x$`x` metres, its velocity is $v$`v` m/s and its acceleration is $a$`a` m/s^{2}.

If $\frac{dx}{dt}=e^{4x}$`d``x``d``t`=`e`4`x`, express $a$`a` in terms of $x$`x`.

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

Apply differentiation methods in solving problems