Kinematics: Question 7

Worked solution

Setting up the problem

Take the direction of motion along the runway as positive throughout. The aircraft starts from rest, so $u=0$, and accelerates uniformly at $a=2.4\text{ m s}^{-2}$.

Part (a): Takeoff speed

Using $v = u + at$ with $u=0$, $a=2.4\text{ m s}^{-2}$, $t=15\text{ s}$: $v = 0 + (2.4)(15)$ $v = 36\text{ m s}^{-1}$

Part (b): Total distance during the $15\text{ s}$

Using $s = ut + \tfrac{1}{2}at^2$ with $u=0$: $s = 0 + \tfrac{1}{2}(2.4)(15)^2 = \tfrac{1}{2}(2.4)(225)$ $s = 270\text{ m}$

As a check, using the average-velocity form $s=\left(\frac{u+v}{2}\right)t = \left(\frac{0+36}{2}\right)(15) = 18\times15=270\text{ m}$, which agrees.

Part (c): Distance travelled during the last $5.0\text{ s}$ of the $15\text{ s}$

The last $5.0\text{ s}$ runs from $t=10\text{ s}$ to $t=15\text{ s}$. To find the distance travelled in this interval, first find the distance travelled during the first $10\text{ s}$, then subtract it from the total distance found in (b).

Distance travelled in the first $10\text{ s}$: $s_{10} = \tfrac{1}{2}(2.4)(10)^2 = \tfrac{1}{2}(2.4)(100) = 120\text{ m}$

Distance travelled in the last $5.0\text{ s}$: $s_{last} = s_{total} - s_{10} = 270 - 120$ $s_{last} = 150\text{ m}$

Check: the velocity at $t=10\text{ s}$ is $v_{10}=u+at=(2.4)(10)=24\text{ m s}^{-1}$. Over the last $5.0\text{ s}$ the aircraft accelerates uniformly from $24\text{ m s}^{-1}$ to $36\text{ m s}^{-1}$, so using the average-velocity form: $s_{last} = \left(\frac{24+36}{2}\right)(5.0) = 30\times5.0 = 150\text{ m}$ This agrees with the subtraction method above.

Final answers

• (a) Takeoff speed $= \boxed{36}\text{ m s}^{-1}$
• (b) Total distance in $15\text{ s}$ $= \boxed{270}\text{ m}$
• (c) Distance in the last $5.0\text{ s}$ $= \boxed{150}\text{ m}$