Question #133012
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?
1
Expert's answer
2020-09-15T10:04:34-0400

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


v0=(16,0) [m/s]\mathbf{v_0} = (16,0) \space [m/s]

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


v1=(0,11.5) [m/s]\mathbf{v_1} = (0, -11.5) \space[m/s]

By definition, the acceleration is:


a=1t[v1v0]=12.3s[(0,11.5)(16,0)]=(16023,5)(6.96,5) [m/s2]\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.


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