a body moving with uniform acceleration travels 84m in First6s and 180m in next 5s . Find the initial velocity and acceleration of body

a = (V2-V1)/T.

V1 = 84m/6s = 14 m/s = Initial velocity.
V2 = 180m/5s = 36 m/s.

T = t1 + t2 = 6 + s = 11s.

To find the initial velocity and acceleration of the body, we can use the kinematic equation for uniformly accelerated motion:

1. First, let's find the acceleration (a) of the body.

The formula for displacement (s) as a function of initial velocity (u), time (t), and acceleration (a) is given by: s = ut + (1/2)at^2

From the first piece of information, we know that the body traveled 84m in the first 6 seconds, so we have:
84 = u(6) + (1/2)a(6)^2

This gives us the equation: 84 = 6u + 18a. ----(1)

2. Next, let's find the initial velocity (u) of the body.

From the second piece of information, we know that the body traveled 180m in the next 5 seconds, so we have:
180 = u(5) + (1/2)a(5)^2

This gives us the equation: 180 = 5u + 12.5a. ----(2)

Now we have two equations with two unknowns (u and a). We can solve them simultaneously to find the values of u and a.

Solving equations (1) and (2) simultaneously:
84 = 6u + 18a ----(1)
180 = 5u + 12.5a ----(2)

Multiplying equation (1) by 5, we get: 420 = 30u + 90a ----(3)

Subtracting equation (2) from equation (3) to eliminate u:
420 - 180 = 30u + 90a - (5u + 12.5a)
240 = 25u + 77.5a ----(4)

Now we have one equation (4) with one unknown (either u or a). To solve for u or a, we need one more equation.

Since we have only two pieces of given information, we can't find the exact values of u and a but we can still find their relationship.

Divide equation (4) by 5 to simplify it:
48 = 5u + 15.5a

Or, we can write it as:
5u + 15.5a = 48 ----(5)

Comparing equation (5) with equation (2), we can see that they are equivalent, which implies that both equations represent the same relationship between u and a.

Therefore, we can conclude that the initial velocity (u) and acceleration (a) of the body cannot be determined uniquely based on the given information.

To find the initial velocity and acceleration of the body, we can use the equations of motion for uniformly accelerated linear motion.

First, let's assume the initial velocity of the body as 'u' and the acceleration as 'a'.

1. Distance traveled in the first 6 seconds:
Using the equation of motion:
s = ut + (1/2)at^2
Here, s = 84m, u = initial velocity, t = 6s, and a = acceleration.
Plugging in the values, we get:
84 = (u * 6) + (1/2)(a * (6)^2) -> Equation 1

2. Distance traveled in the next 5 seconds:
Using the same equation of motion:
s = ut + (1/2)at^2
Here, s = 180m, u = initial velocity, t = 5s, and a = acceleration.
Plugging in the values, we get:
180 = (u * 5) + (1/2)(a * (5)^2) -> Equation 2

Now, we have a system of two equations with two unknowns (u and a). By solving this system of equations, we can find the values of u and a.

Let's solve the equations:
From Equation 1, we can rearrange it as:
84 = 6u + (1/2)(36a)
Multiplying all terms by 2 to get rid of the fraction:
168 = 12u + 36a -> Equation 3

From Equation 2, we can rearrange it as:
180 = 5u + (1/2)(25a)
Multiplying all terms by 2 to get rid of the fraction:
360 = 10u + 25a -> Equation 4

Now, we have two equations (Equation 3 and Equation 4) with two unknowns (u and a). We can solve this system of equations using any suitable method, such as substitution or elimination.

Using the elimination method, multiply Equation 3 by 5 and Equation 4 by 12:
840 = 60u + 180a -> Equation 5
4320 = 120u + 300a -> Equation 6

Subtract Equation 5 from Equation 6:
4320 - 840 = 120u - 60u + 300a - 180a
3480 = 60u + 120a

Divide both sides by 60:
3480/60 = (60u + 120a)/60
58 = u + 2a

Now, we have found the value of "u + 2a" as 58.

Next, let's substitute this value back into Equation 5 or Equation 6 to solve for 'u':
840 = 60u + 180a -> Equation 5

Substituting 58 for 'u + 2a' in Equation 5:
840 = 60(58 - 2a) + 180a
840 = 3480 - 120a + 180a
840 = 3480 + 60a

Rearranging the terms:
60a = 3480 - 840
60a = 2640

Dividing both sides by 60:
a = 2640/60
a = 44 m/s^2

Finally, we have found the value of the acceleration 'a' as 44 m/s^2.

To find the initial velocity 'u', we can substitute this value back into Equation 3 or Equation 4:
180 = 5u + (1/2)(25 * 44)
180 = 5u + 11 * 44
180 = 5u + 484
5u = 180 - 484
5u = -304

Dividing both sides by 5:
u = -304/5
u = -60.8 m/s

Therefore, the initial velocity 'u' is approximately -60.8 m/s, and the acceleration 'a' is 44 m/s^2.