Question

a body covers 20m in 7th second and 24m in the 9th second how much distance it covers in 15th second

Answer 1

Answer:

• 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 m

$hope \: its \: help \: u$

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