I've asked this question several times before
please answer it correct this time
An aeroplane taking off from a field has a run of 500 m. What is the acceleration and take off velocity if it leaves the ground 10 seconds after the start ?
Answers
Explanation:
What is the acceleration and take off velocity if it leaves the ground 10 seconds after the start ? The length of the runway is 500 m . The aeroplane leaves the ground 10 seconds after the start . => ( 500 m / 10 s ) = 50 m/s .
Answer:
Distance travelled, s = 500 m
♦ Initial velocity, u = 0 m/s 〘As the aeroplane starts from rest〙
♦ Time, t = 10 s
TO FinD :-
The acceleration and the take off velocity of the aeroplane.
AcknowledgemenT :-
♦ s = ut + 1/2 at²
\textsl{Where,}\textslWhere,
s is the distance travelled.
u is the initial velocity
t is the time taken
a is the acceleration
♦ v = u + at
\textsl{Where,}\textslWhere,
v is the final velocity.
u is the initial velocity
a is the acceleration
t is the time taken
SolutioN :-
\sf{s = ut + \dfrac{1}{2} at\²}s=ut+
2
1
at\²
\textsl{Putting the values :-}\textslPuttingthevalues:−
\begin{gathered}\sf{\implies 500=0(10)+\dfrac{1}{2}a(10)^2}\\\\\sf{\implies 500=0+\dfrac{100a}{2}}\\\\\sf{\implies 500=50a}\\\\\sf{\implies a=\dfrac{500}{50}}\\\\\underline{\boxed{\sf{\implies a=10\ m/s^2}}}\end{gathered}
⟹500=0(10)+
2
1
a(10)
2
⟹500=0+
2
100a
⟹500=50a
⟹a=
50
500
⟹a=10 m/s
2
\rule{150}{2}
\sf{v=u+at}v=u+at
\textsl{Putting the values :-}\textslPuttingthevalues:−
\begin{gathered}\sf{\implies v=0+(10)(10)}\\\\\sf{\implies v=(10)^2}\\\\\underline{\boxed{\sf{\implies v=100\ m/s}}}\end{gathered}
⟹v=0+(10)(10)
⟹v=(10)
2
⟹v=100 m/s
Therefore, the acceleration and the take off velocity of the aeroplane are 10 m/s² and 100 m/s respectively.