To solve this problem, we can use the kinematic equation that relates displacement, initial velocity, time, and acceleration:
x = ut + (1/2)at^2
where x is the displacement, u is the initial velocity, t is the time, and a is the acceleration.
Given that the body has a uniform acceleration, we can assume the acceleration remains constant throughout the motion.
Let's break down the problem into two parts:
Part 1: At t = 0, the x-coordinate is 300m, and the velocity is 10m/s. Using the equation above, we have:
300 = (10)(0) + (1/2)a(0)^2 300 = 0
This equation doesn't provide any useful information about the acceleration.
Part 2: At t = 4 seconds, the x-coordinate is -50m. Using the equation above, we have:
-50 = (10)(4) + (1/2)a(4)^2 -50 = 40 + 8a
Rearranging the equation, we get:
8a = -50 - 40 8a = -90 a = -90/8 a = -11.25 m/s^2
Therefore, the magnitude of the acceleration is 11.25 m/s^2.
Note: In this problem, the negative sign indicates that the body is decelerating or changing direction.