Question

A body moving with uniform acceleration cover 20m in 7th second and 24 m in 9th second of its motion find the distance it will cover in the 15 second of its motion

Answer 1

answer is
S = ut + (at^2)/2

20 = u.1 + (a.1^2)/2

20 = u + a /2 —— (1)

For 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+2at) + (a.1^2 )/2

24 = u + 2a.1 + a/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

For 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

QED.