A body travel a distance of 20m in the 7th second and 24m in 9th second. How much distance shall it travel in the 15th second.
Answers
αηѕωєя
36 Meters
Explaηation
Assuming uniform acceleration a (without assuming this, infinite number of solutions is possible), and initial speed u at 7th second, the distance travelled:
S = ut + (at^2)/2
20 = u.1 + (a.1^2)/2
20 = u + a /2 —————— (1)
At 9th second, initial speed v would be :
v = u + a(t2 - t1 ) = u + a (9–7) = u + 2a ——— (2)
Now, distance for 9th second :
S = vt + (at^2)/2
Using (2) and substituting given values:
24 = (u+2a)t + (a.t^2 )/2
24 = (u + 2a).1 + (a.1)/2 = u + 5a/2 ——— (3)
Subtracting eq. (1) from eq. (3), we get :
4 = 2a, or a = 2 m/s^2
Substituting this back in (1), we get:
u = 19 m/s
At 15th second, initial speed w is given by:
w = u + at,
where u is initial speed of 7th second and t = 15 - 7 = 8. Hence,
w = 19 + 2*8 = 35 m/s
Distance s travelled in 15th second:
S = wt + (at^2)/2 = 35*1 + (2*1^2)/2 = 36 mQED.
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