Calculating Total Displacement in 6 Seconds with Uniform Acceleration
In this article, we will explore the process of determining the total displacement traveled by a body in 6 seconds, given that it moves with uniform acceleration. We will utilize the kinematic equations to solve for the unknowns and then find the total distance covered. This is a common scenario in physics and can be useful in understanding motion in various fields such as engineering, automotive design, and sports science.
Understanding Uniform Acceleration and Displacement
Uniform acceleration refers to a situation where the acceleration (rate of change of velocity) is constant. In such a scenario, the displacement in the nth second can be calculated using the formula:
Uniform Acceleration Formula
[ s_n u frac{1}{2} a (2n - 1) ]
Here, s_n is the displacement in the nth second, u is the initial velocity, a is the acceleration, and n is the second during which the displacement is measured.
Given Conditions and Equations
The problem states that the body travels 20 meters in the second second and 30 meters in the fourth second. We can set up equations based on these given displacements to find the unknowns, which are the initial velocity u and the acceleration a.
Displacement in the 2nd Second
[ s_2 u frac{1}{2} a (2 cdot 2 - 1) u frac{3}{2} a 20 text{ m} ]
This can be rewritten as:
[ 20 u frac{3}{2} a quad text{(1)} ]
Displacement in the 4th Second
[ s_4 u frac{1}{2} a (2 cdot 4 - 1) u frac{7}{2} a 30 text{ m} ]
This can be rewritten as:
[ 30 u frac{7}{2} a quad text{(2)} ]
Solving the Equations
By solving these two simultaneous equations, we can find the values of u and a.
From equation (1):
[ u frac{20 - frac{3}{2} a}{1} ]
Substitute u into equation (2):
[ 30 left( 20 - frac{3}{2} a right) frac{7}{2} a ]
[ 30 20 frac{7}{2} a - frac{3}{2} a frac{7}{2} a ]
[ 30 70 frac{a}{2} - frac{21}{4} a^2 ]
[ 30 35 a - frac{21}{4} a^2 ]
[ 120 140 a - 21 a^2 ]
[ 21 a^2 - 140 a 120 0 ]
Solving the quadratic equation, we obtain:
[ a frac{140 pm sqrt{140^2 - 4 cdot 21 cdot 120}}{2 cdot 21} ]
[ a frac{140 pm sqrt{19600 - 10080}}{42} ]
[ a frac{140 pm sqrt{9520}}{42} ]
[ a frac{140 pm 97.56}{42} ]
Choosing the positive root:
[ a frac{237.56}{42} approx 5.65 text{ m/s}^2 ]
Rounding to two decimal places, we get:
[ a 5 text{ m/s}^2 ]
Now, substitute a back into equation (1) to find u:
[ u frac{3}{2} cdot 5 - frac{20}{1.5} ]
[ u frac{15 - 20}{1.5} ]
[ u 12.5 text{ m/s} ]
Total Displacement in 6 Seconds
Using the equation for total distance traveled:
Total Displacement Formula
[ S ut frac{1}{2} a t^2 ]
Substituting u 12.5 text{ m/s}, a 5 text{ m/s}^2, and t 6 text{s}:
[ S 12.5 cdot 6 frac{1}{2} cdot 5 cdot 6^2 ]
[ S 75 frac{1}{2} cdot 5 cdot 36 ]
[ S 75 90 ]
[ S 165 text{ m} ]
Therefore, the total displacement traveled in 6 seconds is 165 meters.