A uniformly accelerating body covers 18m in 9th second and 20m it 11th second. How far will it
travel in the next 5 seconds?
Answers
Answer:
We know that distance Travelled by uniformly accelerated body in n
th e s is given by
Sn =u+ 1/2 a(2n -1) where a is the acceleration of the body and u is the initial velocity
it is given that Sn =18m at n=9
18 = u + 17a/2------------>eqn(1)
it is given that sn = 20m at n = 11
20 = u + 21a/2--------------> eqn (2)
solve eqn 1 and 2 we get
18-17a/2 = u
20 - 21a/2 =u
18-17a/2 = 20 -21a/2
21a/2 - 17a/2 = 20 -18
4a/2 = 2
2a = 2
a = 1 m/s
solving for u
18 - 17(1)/2 = u
u = 36 -17/2
u = 9.5 m/s
Now we have find velocity at t = 11s
v = u +at
= 9.5 + 1(11)
= 11+9.5
= 20.5 m/s
Now distance during next 5 s can be calculated ,
s = ut + 1/2 at²
=20.5×5 +1/2 ×1×5² now u=20.5m/s
= 102.5+12.5
= 115m
Hence 115m is the required answer
Hope it will clear your doubt