Respuesta :
Answer:
15 seconds.
Step-by-step explanation:
∵ The distance covered by plane in first second = 100 ft,
Also, in each succeeding second it climbs 100 feet more than it climbed during the previous second,
So, distance covered in second second = 200,
In third second = 300,
In fourth second = 400,
............, so on....
Thus, the total distance covered by plane in n seconds = 100 + 200 + 300 +400......... upto n seconds
[tex]=\frac{n}{2}(2\times 100 + (n-1)100)[/tex] ( Sum of AP )
[tex]=\frac{n}{2}(200+100n-100)[/tex]
[tex]=\frac{n}{2}(100+100n)[/tex]
[tex]=50n+50n^2[/tex]
Suppose the distance covered in n seconds is 12,000 feet,
[tex]\implies 50n+50n^2=12000[/tex]
[tex]n+n^2=240[/tex]
[tex]n^2+n-240=0[/tex]
[tex]n^2+16n-15n-240=0[/tex]
[tex]n(n+16)-15(n+16)=0[/tex]
[tex](n-15)(n+16)=0[/tex]
[tex]\implies n=15\text{ or }n=-16[/tex]
∵ n can not be negative,
Hence, after 15 seconds the plane will reach an altitude of 12,000 feet above its takeoff height.
Answer:
Step-by-step explanation:
After t seconds, the airplane's altitude (in feet) is 100 + 200 + . . . + 100t = 100(1 + 2 + . . . + t) = 100*t(t + 1)/2 = 50t(t + 1). Thus, we want to find the smallest t such that 50t(t + 1) [tex]\geq[/tex] 12000. Dividing both sides by 50, we get t(t + 1) [tex]\geq[/tex] 240. Since $15*16 = 240$, the smallest such t is t = 15.
