An object moving with a constant accleration travels distance of 20m and 24m respectively in 3rd and 4th second of its motion. Find the initial velocity and the accleration of an object
To find the initial velocity and acceleration of the object, we can use the equations of motion.
Let's assume:
- The initial velocity of the object is u.
- The acceleration of the object is a.
- The distance traveled by the object in the 3rd second is s1 = 20m.
- The distance traveled by the object in the 4th second is s2 = 24m.
We know that the distance traveled by an object with constant acceleration can be calculated using the equation:
s = ut + (1/2)at^2
Using this equation, we can find the relationship between the distance traveled and the time taken for an object with constant acceleration.
For the 3rd second:
s1 = u(3) + (1/2)a(3^2)
20 = 3u + 4.5a ...(Equation 1)
For the 4th second:
s2 = u(4) + (1/2)a(4^2)
24 = 4u + 8a ...(Equation 2)
We now have a system of two equations (Equation 1 and Equation 2) that we can solve simultaneously to find the values of u (initial velocity) and a (acceleration) of the object.
To solve this system of equations, we can use a variety of methods such as substitution, elimination, or matrices. Let's use the substitution method.
Rearrange Equation 1 for u:
20 - 4.5a = 3u
u = (20 - 4.5a)/3 ...(Equation 3)
Substitute Equation 3 into Equation 2:
24 = 4((20 - 4.5a)/3) + 8a
24 = (80 - 18a)/3 + 8a
Multiply both sides by 3 to eliminate the fraction:
72 = 80 - 18a + 24a
72 = 80 + 6a
Rearrange the equation:
6a = 80 - 72
6a = 8
a = 8/6
a = 4/3
Substitute the value of a back into Equation 3 to find u:
u = (20 - 4.5(4/3))/3
u = (20 - 6)/3
u = 14/3
Therefore, the initial velocity (u) is 14/3 m/s and the acceleration (a) is 4/3 m/s².