Question

Example 2: An aeroplane taxies on the runway for 30 s with an acceleration of 3.2 m/s
before taking off. How much distance would it have covered on the runway?​

Answer 1

Answer:

The aeroplane would have covered 1440 m on the runway.

Explanation:

Given :-

• An aeroplane taxies on the runway for 30 s with an acceleration of 3.2 m/s² before taking off.

To find :-

• How much distance would it have covered on the runway ?

Solution :-

Formula used :

${\boxed{\sf{S=ut+\dfrac{1}{2}at^2}}}$

• t = time = 30 s
• a = acceleration = 3.2 m/s²
• u = initial velocity = 0 m/s

[put values]

$\mapsto\sf{S=0\times\:t+\dfrac{1}{2}\times\:3.2\times\:(30)^2}$

$\mapsto\sf{S=0+(16\times\:3\times\:30)}$

$\mapsto\sf{S=1440}$

Therefore, it would have covered 1440 m on the runway.

____________________

Some formulas :

• $\sf{v=u+at}$

• $\sf{v^2=u^2+2as}$

• $\sf{s_n=u+\dfrac{1}{2}a(2n-1)}$

Answer 2

Answer:

• In this question, we have to find the distance covered by aeroplane taxis on the runway for calculating distance we have to use second equation of motion :]

$\bigstar\:\: \underline{\overline{\boxed{ \sf s = ut + \dfrac{1}{2}at^2}}} \: \: \bigstar$

$\tt {\pink{Here}}\begin{cases} \sf{\green{Acceleration \: (a)=3.2 \: m/s^2}}\\ \sf{\blue{Initial \: Velocity \: (u)=0 \: m/s}}\\ \sf{\orange{Time \: (t) =30 \: s}}\\ \sf{\red{Distance \: (s)=?}}\end{cases}$

$:\implies \sf s = 0 \times 30 + \dfrac{1}{2} \times 3.2 \times (30)^2$

$:\implies \sf s = 0 + \dfrac{1}{2} \times 3.2 \times 900$

$:\implies \sf s = \dfrac{1}{2} \times 3.2 \times 900$

$:\implies \sf s = 450\times 3.2$

$:\implies \underline{ \boxed{\sf s = 1440 \: m }}$

$\therefore\underline{\textsf{Distance covered by aeroplane taxis on the runway is \textbf{ 1440 m}}}.$