Question

A body is moving with uniform acceleration and it covers 20m in 4th second and 30m in 8th second. Find (i) the initial velocity, (ii) the acceleration of the body.​

Answer 1

$s = ut + \frac{1}{2} {at}^{2}$

$20 = u(4) + \frac{1}{2} a{(4)}^{2}$

$20 = 4u + \frac{4a}{2}$

$4u = 20 - \frac{4a}{2}$

$S = 20+ 30 = 50$

$t = 4 + 8 = 12$

$50 = 12(u) + \frac{1}{2} a( {12)}^{2}$

$50 = 12u + 72a$

$50 = 12(20 - \frac{4a}{2} ) + 72a$

$50 = 240a - 24a + 72a$

$50 = 240a + 48a$

$50 = 288a$

$a = \frac{288}{50}$

$a = 5.76 \: m {s}^{2}$

Answer 2

Answer:

Let initial velocity u and acceleration is a.

For the first part of motion traveled 84 m in 6 seconds. By equation,

s = ut + 1/2 at square

The above kinetic equation gives the following equation

6u + 18a = 84 …..(1)

The final velocity at the end of 6 seconds is

u6 = u+6a, where u6 is velocity after 6 seconds

After using velocity relation after 6 seconds in the second part of the motion we get the following equation

(u+6a)5 + ½× a× 25 = 180

By simplification we will get :- 5u + (85/2)a = 180……..(2)

By solving equations (1) and (2) we get

a = 2 m/s2 , u = 4 m/s