an object travels in a straight line at a constant acceleration for 6 seconds.in the first 2 seconds it travels 18 metres and in entire 6 seconds it covers 102 metres. find the acceleration.
102=vi(6)+1/2 a 36
18=vi(2)+ 1/2 a 4
multiply the second equation by 3, then subtract the second equation from the first.
d = Vi t + (1/2) a t^2
18 = Vi (2) + (1/2) a (4)
102 = Vi (6) + (1/2) a (36)
18 = 2 Vi + 2 a
or
9 = Vi + a
102 = 6 Vi + 18 a
or
17 = Vi + 3 a
8 = 2 a
a = 4 m/s^2
and
Vi = 5 m/s
thnx guyz !!!!!!!!
To find the acceleration of the object, we can use the equations of motion.
Let's assume the initial velocity of the object is "u" (m/s), the acceleration is "a" (m/s²), and the time taken is "t" (s).
We are given two pieces of information:
1) In the first 2 seconds, the object travels 18 meters.
Using the equation of motion:
s = ut + (1/2)at²
18 = u(2) + (1/2)a(2)²
18 = 2u + 2a
2) In the entire 6 seconds, the object covers 102 meters.
Again using the equation of motion:
s = ut + (1/2)at²
102 = u(6) + (1/2)a(6)²
102 = 6u + 18a
Now we have two equations with two unknowns (u and a).
Let's solve these equations simultaneously to find the values of u and a.
First, we'll rearrange both equations to solve for u in terms of a:
Equation 1: 18 = 2u + 2a
=> 2u = 18 - 2a
=> u = 9 - a (Equation 3)
Equation 2: 102 = 6u + 18a
=> 6u = 102 - 18a
=> u = 17 - 3a (Equation 4)
Since both Equation 3 and Equation 4 are equal to "u", we can set them equal to each other:
9 - a = 17 - 3a
4a = 8
a = 2
Therefore, the acceleration of the object is 2 m/s².