Answer to Question #133012 in Physics for Jade

Question

A particle is initially moving along the positive x-axis at a speed of 16.0m/s. After 2.30s the particle is moving along the negative y-axis at a speed of 11.5m/s. Find the x and y components of the particles acceleration.
What is the particles acceleration in the x direction?
What is the particles acceleration in the y direction?

Answer 1

The vector of the initial speed (along x-axis) has the following coordinates:

$$\mathbf{v_0} = (16,0) \space [m/s]$$

The vector of the final speed (along y-axis) has the following coordinates:

$$\mathbf{v_1} = (0, -11.5) \space[m/s]$$

By definition, the acceleration is:

$$\mathbf{a} = \dfrac{1}{t}[\mathbf{v_1} - \mathbf{v_0}] = \dfrac{1}{2.3s}[(0,-11.5) - (16,0)] = \left( -\dfrac{160}{23}, -5\right)\approx (-6.96,-5) \space [m/s^2]$$

Answer. x-direction: -6.96 m/s^2, y-direction: -5 m/s^2.