To solve this problem, we can use the equations of motion for uniformly accelerated motion. One such equation relates displacement (Δx), initial velocity (v0), time (t), and acceleration (a):
Δx = v0 * t + (1/2) * a * t^2
Given that the body's initial velocity (v0) is 10 m/s, its initial displacement (x0) is 300 m, and the displacement after 4 seconds (x) is -50 m, we can write two equations:
For the initial position: x0 = v0 * t + (1/2) * a * t^2 ---- (Equation 1)
For the position after 4 seconds: x = v0 * (t + 4) + (1/2) * a * (t + 4)^2 ---- (Equation 2)
Substituting the values into the equations, we get: Equation 1: 300 = 10 * t + (1/2) * a * t^2 Equation 2: -50 = 10 * (t + 4) + (1/2) * a * (t + 4)^2
Now, we can solve these two equations simultaneously to find the values of 'a' and 't'. Once we have the value of 'a', we can determine its magnitude.
Simplifying Equation 1, we get: 300 = 10t + (1/2)at^2
Simplifying Equation 2, we get: -50 = 10t + 40 + (1/2)at^2 + 4at + 8t
Combining like terms in Equation 2, we have: -50 = (1/2)at^2 + (12 + 8a)t + 40
Now, we have a system of two equations:
- 300 = 10t + (1/2)at^2
- -50 = (1/2)at^2 + (12 + 8a)t + 40
We can solve this system using algebraic methods or numerical methods such as substitution or graphical analysis to find the values of 'a' and 't'. Once we determine the value of 'a', we can find its magnitude.