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? with full answer
Answers
Answered by
6
Answer:brainliest pls
Time(t) = 30 s. s = 1440 m. Total distance covered (s) = 1440 metres.
Explanation:
Initial speed of airplane u=0 m/s
Acceleration of the plane a=3.2 m/s
2
and time taken t=3.28 s
Using formula, distance covered S=ut+
2
1
at
2
⇒ S=0(32.8)+0.5(3.2)(32.8)
2
=1721.3∼1720m
Answered by
3
Answer:
The aeroplane would have covered 1440 m distance.
Explanation:
a = 3.2 m/s²
t = 30 s
u = 0 m/s
s = ut + ½ at²
=> s = 0 * 30 + ½ * 3.2 * 30²
=> s = 0 + ½ * 3.2 * 900
=> s = ½ * 900 * 3.2
=> s = 450 * 3.2
=> s = 1440 m
The aeroplane would have covered 1440 m distance.
Similar questions