Question

A body covers a distance of 20m in the 7th second and 24m in the 9th second . how much shall it cover in 15th second.

Answer 1

36m in 15th second
how to solve this...
use the 2nd equation of motion...to find U(intial velocity) and A(acceleration)

It is given Distance covered by an object in 7th second. Which will be equal to( Distance covered by Object in 7sec) - (Distance covered by Object in 6sec ) = 20M

After solving the above you will be left with two variable U and A

just use the 2nd Case to determine U and A.which is distance travelled by body in 9th second. Now you have two equation and Two variable which can be easily find out.

After getting U and A you can find distance covered by Object in 15th second by
(Distance covered by body in 15sec) - (Distance covered by body in 14 sec)
which will give you 36

If you didn't understand comment I will send you a solution in copy.

Now one more thing ,
Distance covered by an object in Nth sec is given by S = u+a(2n-1)/2

Answer 2

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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