 New Zealand
Level 8 - NCEA Level 3

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 $\frac{\mathrm{d}y}{\mathrm{d}t}$dydt and also $\frac{\mathrm{d}x}{\mathrm{d}t}$dxdt and then find $\frac{\mathrm{d}y}{\mathrm{d}x}$dydx from the relation

$\frac{\mathrm{d}y}{\mathrm{d}t}=\frac{\mathrm{d}y}{\mathrm{d}x}.\frac{\mathrm{d}x}{\mathrm{d}t}$dydt=dydx.dxdt

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)=dxdt and

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

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)=dvdt=dvdx.dxdt

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

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

Example

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=loge x. Express the acceleration as a function of $x$x.

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

Worked Examples

Question 1

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/s2.

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

Question 2

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/s2.

If $\frac{dx}{dt}=e^{4x}$dxdt=e4x, express $a$a in terms of $x$x.

Outcomes

M8-11

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

91578

Apply differentiation methods in solving problems