Question

An experiment on the take off performance of an aeroplane shows that the acceleration varies as shown in and that it takes $12s$ to take off from a rest position.
a. Write the acceleration vs time, velocity vs time and position vs time relations for completer journey.
b. Plot velocity vs time relation for the motion.
c. Find the distance along the run away covered by the aeroplane.

Answer 1

For time $t=0$ to $6s$
We can write equation for this motion, it is $y=mx$ type line where $m$ is the slope of the line.
Here $a=\frac{(5-0)}{(6-0)}\cdot t\Rightarrow a=\frac{5}{6}t\left(m{s}^{-2}\right)....\left(i\right)$
(For $0\le t<6s$)
From $6s$ to $12s$ acceleration is constant. we can write equation for this motion, it is $y=constant$.
Hence $a=5\left(m{s}^{-2}\right)$ (For $6\le t<12s)$
For time $t=0$ to $6s$
From Eq. (i), $a=\frac{dv}{dt}=\frac{5}{6}t\Rightarrow dv=\frac{5}{6}tdt$
Integrating both sides, we get ${\int }_0^{v}dv=\frac{5}{6}{\int }_0^{t}tdt$
$[v{]}_0^v=\frac{5}{6}{\left[\frac{t^2}{2}\right]}_0^t$ or $(v-0)=\frac{5}{12}(t^2-0)$ or $v=\frac{5}{12}{t}^2$
Here velocity at $t=6s$ is $v_{(t=6s)}=\frac{5}{12}\times 6^2=15m{s}^{-1}$
For time $t=0$ to $6s$
The acceleration is constant.
$a=5\left(m{s}^{-2}\right)$ or $\frac{dv}{dt}=5$ or $dv=5dt$
Integrating both sides, ${\int }_{15}^v={\int }_6^{t}5\cdot dt$
$[v{]}_{15}^v=5[t{]}_6^t$
$v-15=5[t-6]$
$\Rightarrow v=5t-15$
Velocity - time relation will be a straight line of type $y=mx+c$.
Plotting velocity - time relations

For time $t=0$ to $6s$ $v=\frac{5}{12}{t}^2$ Velocity time graph will be parabola passing through origin.

For time $t=6s$ to $12s$ $v-5t-15$ At $t=12s$ $v=5\left(12\right)-15=45m{s}^{-1}$

Velocity time graph of complete motion

Position time relations
For time $t=0$ to $ts$
As we know $v=\frac{dx}{dt}=\frac{5}{12}{t}^2$
Hence, $\frac{5}{12}{t}^2=\frac{dx}{dt}\Rightarrow \frac{5}{12}\int t^{2}dt$
Integrating both sides, ${\int }_0^{x}dx=\frac{5}{12}{\int }_0^t{t}^{2}dt$
$(x-0)=\frac{5}{12}{\left[\frac{t^3}{3}\right]}_0^t$
$\Rightarrow x=\frac{5}{36}{t}^3$
Position or the distance travelled from $0$ to $6s$
$x_{(t=6\sec )}=\frac{5}{36}\times (6{)}^3=30m$
For time $t=6s$ to $12s$
$v=5t-15\Rightarrow \frac{dx}{dt}=5t-5$
or $dx=(5t-15)dt$
Integrating both sides
${\int }_{30}^{x}dx={\int }_6^t(5t-15)dt$
$[x{]}_{30}^x=5{\int }_0^{t}tdt-15{\int }_6^{t}dt$
$(x-30)=5{\left[\frac{t^2}{2}\right]}_6^t-15[t{]}_6^t$
$x-30=\frac{5}{2}[t^2-6^2]-15[t-6]$
$x=\frac{5}{2}{t}^2-15t+30$
At $t=12s$,
$x_{(t=12s)}=\frac{5}{2}(12{)}^2-15\times 12+30=210m$.