Question

An experiment on the take off performance of an aeroplane shows that the acceleration vaies as shown in (figure) 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 complete journey.
b. Plot velocity vs. time relation for the motion.
c. Find the distance along the run way 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=((5-0))/((6-0))*timplies a=5/6t(ms^-2)` …(i)
`(For 0letlt6s)`
From `6s` to `12s` acceleration is constant, we can write equation for this motion, it is `y=const ant`.
Hence `a=5(ms^-2)` `(For 6letlt12s)`
For time `t=0` to `6s`
From Eq. (i) `a=(dv)/(dt)=5/6timpliesdv=5/6tdt`
Integrating both sides, we get `underset0oversetvintdv=5/6underset0oversettinttdt`
`[v]_0^v=5/6[t^2/2]_0^t` or `(v-0)=5/12(t^2-0)` or `v=5/12t^2`
Here velocity at `t=6s` is `v_(t=6s)=5/12xx6^2=15ms^-1`
For time `t=0` to `6s`
The acceleration is constant.
`a=5(ms^-2)` or `(dv)/(dt)=5` or `dv=5dt`
Integrating both sides, `underset15oversetvintdv=underset6oversettint5*dt`
`[v]_15^v=5[t]_6^t`
`v-15=5[t-6]`
`impliesv=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=5/12t^2`
Velocity time graph will be parabola passing through origin.

Position time relations
For time `t=0` to `6s`
As we know `v=(dx)/(dt)=5/12t^2`
Hence, `5/12t^2=(dx)/(dt)impliesdx=5/12int t^2dt`
Integrating both sides, `underset (0) oversetx int dx=5/12underset (0) overset (t) int t^2dt`
`(x-0)=5/12[t^3/3]_0^timpliesx=5/36t^3`
Position or the distance travelled from 0 to 6s
`x_((t=6sec))=5/36xx(6)^3=30m`
For time `t=6s` to `12s`
`v=5t-15implies(dx)/(dt)=5t-5`
or `dx=(5t-15)dt`
Integrating both sides
`underset(30)overset(x)intdx=underset6oversettint(5t-15)dt`
`[x]_30^x=5underset6oversettinttdt-15underset6oversettintdt`
`(x-30)=5[t^2/2]_6^t-15[t]_6^t`
`x-30=5/2[t^2-6^2]-15[t-6]`
`x=5/2t^2-15t+30`
At `t=12s`,
`x_((t=12s))=5/2(12)^2-15xx12+30=210m`