A body moving with uniform acceleration travels 84mts in the first 6 second and180m in the next 5 sec. Find the initial velocity & acceleration of bodyA) 23 / s , 4 /m m s B) 29 / s, 2 / sm m C) 22 / s , 3 /m m s D) 22 / s , 4 /m m s

To solve this problem, we need to use the equations of motion.

1. For the first 6 seconds, we have the equation: s = ut + 0.5at^2 where s is the distance covered, u is the initial velocity, a is the acceleration, and t is the time. Substituting the given values, we get: 84 = 6u + 0.5a(6)^2 This simplifies to: 84 = 6u + 18a ----(1)

2. For the next 5 seconds, the initial velocity is the final velocity of the first 6 seconds. So, the equation becomes: s = ut + 0.5at^2 The distance covered in the next 5 seconds is 180 - 84 = 96m. Substituting the values, we get: 96 = (u + 6a)5 + 0.5a(5)^2 This simplifies to: 96 = 5u + 30a + 12.5a 96 = 5u + 42.5a ----(2)

3. Now, we have two equations (1 and 2) and two unknowns (u and a). We can solve these equations simultaneously to find the values of u and a.

By solving these equations, we find that the initial velocity (u) is 4 m/s and the acceleration (a) is 2 m/s^2.

So, the correct answer is B) 4 m/s, 2 m/s^2.

To solve this problem, we need to use the equations of motion.

1. For the first 6 seconds, we have the equation: s = ut + 0.5at^2 where s is the distance covered, u is the initial velocity, a is the acceleration, and t is the time. Substituting the given values, we get: 84 = 6u + 0.5a(6)^2 Simplifying this, we get: 84 = 6u + 18a ----(1)

2. For the next 5 seconds, the initial velocity will be the final velocity of the first 6 seconds. So, the equation becomes: s = ut + 0.5at^2 But u here is u + 6a (from the first equation of motion, v = u + at) So, the equation becomes: 180 = (u + 6a)5 + 0.5a(5)^2 Simplifying this, we get: 180 = 5u + 30a + 12.5a 180 = 5u + 42.5a ----(2)

3. Now, we have two equations (1 and 2) and two unknowns (u and a). We can solve these equations simultaneously to find the values of u and a.

   Subtract equation (1) from equation (2), we get: 96 = -u + 24.5a Rearranging, we get: u = -96 + 24.5a ----(3)

   Substituting equation (3) into equation (1), we get: 84 = 6(-96 + 24.5a) + 18a Simplifying, we get: 84 = -576 + 147a + 18a 84 = -576 + 165a Adding 576 to both sides, we get: 660 = 165a Dividing by 165, we get: a = 4 m/s^2

   Substituting a = 4 into equation (3), we get: u = -96 + 24.5*4 u = -96 + 98 u = 2 m/s

So, the initial velocity (u) is 2 m/s and the acceleration (a) is 4 m/s^2. Therefore, the correct answer is B) 2 m/s, 4 m/s^2.