A boy starts with initial speed u and mouse in a straight line with constant acceleration it has been be after covering a distance 2S its speed just after covering distance s is
speed u and mouse ?
it has been be after?
Makes no sense to me.
To find the speed of the boy just after covering a distance s, we need to use the equations of motion along with the given information.
Let's break down the problem step by step:
1. Let's assume the initial speed of the boy is u and the acceleration is a.
2. We know that the distance covered by the boy is directly proportional to the square of time, which can be expressed as: s = ut + (1/2)at^2 (equation 1).
3. Given that the boy covers a distance of 2S, we can modify equation 1 to: 2S = u(2t) + (1/2)a(2t)^2.
4. Simplifying the equation, we have: 2S = 2ut + 2at^2.
5. Dividing the equation by 2, we get: S = ut + at^2 (equation 2).
6. Now, we want to find the speed of the boy just after covering a distance s, which is equivalent to finding the velocity (speed) v at that particular time t.
7. We can relate the velocity v with the initial velocity u and the acceleration a using the equation v = u + at (equation 3).
8. Rearrange equation 2 to express time t: t = (S - ut) / at.
9. Substitute the value of t into equation 3: v = u + a((S - ut) / at).
10. Cancel out the 'at' terms: v = u + (S - ut).
11. Simplify the equation: v = u + S - ut.
12. Rearrange the equation: v = u(1 - t) + S.
13. The speed of the boy just after covering distance s is given by the equation: v = u(1 - t) + S.
By substituting the values of u, a, and S into the equation, you can calculate the speed v of the boy just after covering a distance s.