Question

A skateboarder travels on a horizontal surface with an initial velocity of 3.6 m/s toward the south and a constant acceleration of 1.8 m/s^2 toward the east. Let the x direction be eastward and the y direction be northward, and let the skateboarder be at the origin at t=0.

a. What is her x position at t=0.60s?
b. What is her y position at t=0.60s?
c. What is her x velocity component at t=0.60s?
d. What is her y velocity component at t=0.60s?

Answer 1

Answer:

a) The x-position of the skateboarder is 0.324 meters.

b) The y-position of the skateboarder is -2.16 meters.

c) The x-velocity of the skateboard is 1.08 meters per second.

d) The y-velocity of the skateboard is -3.6 meters per second.

Explanation:

a) The x-position of the skateboarder is determined by the following expression:

$x(t) = x_{o} + v_{o,x}\cdot t + \frac{1}{2}\cdot a_{x} \cdot t^{2}$ (1)

Where:

$x_{o}$ - Initial x-position, in meters.

$v_{o,x}$ - Initial x-velocity, in meters per second.

$t$ - Time, in seconds.

$a_{x}$ - x-acceleration, in meters per second.

If we know that $x_{o} = 0\,m$, $v_{o,x} = 0\,\frac{m}{s}$, $t = 0.60\,s$ and $a_{x} = 1.8\,\frac{m}{s^{2}}$, then the x-position of the skateboarder is:

$x(t) = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(1.8\,\frac{m}{s^{2}} \right) \cdot (0.60\,s)^{2}$

$x(t) = 0.324\,m$

The x-position of the skateboarder is 0.324 meters.

b) The y-position of the skateboarder is determined by the following expression:

$y(t) = y_{o} + v_{o,y}\cdot t + \frac{1}{2}\cdot a_{y} \cdot t^{2}$ (2)

Where:

$y_{o}$ - Initial y-position, in meters.

$v_{o,y}$ - Initial y-velocity, in meters per second.

$t$ - Time, in seconds.

$a_{y}$ - y-acceleration, in meters per second.

If we know that $y_{o} = 0\,m$, $v_{o,y} = -3.6\,\frac{m}{s}$, $t = 0.60\,s$ and $a_{y} = 0\,\frac{m}{s^{2}}$, then the x-position of the skateboarder is:

$y(t) = 0\,m + \left(-3.6\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(0\,\frac{m}{s^{2}}\right)\cdot (0.60\,s)^{2}$

$y(t) = -2.16\,m$

The y-position of the skateboarder is -2.16 meters.

c) The x-velocity of the skateboarder ($v_{x}$), in meters per second, is calculated by this kinematic formula:

$v_{x}(t) = v_{o,x} + a_{x}\cdot t$ (3)

If we know that $v_{o,x} = 0\,\frac{m}{s}$, $t = 0.60\,s$ and $a_{x} = 1.8\,\frac{m}{s^{2}}$, then the x-velocity of the skateboarder is:

$v_{x}(t) = \left(0\,\frac{m}{s} \right) + \left(1.8\,\frac{m}{s} \right)\cdot (0.60\,s)$

$v_{x}(t) = 1.08\,\frac{m}{s}$

The x-velocity of the skateboard is 1.08 meters per second.

d) As the skateboarder has a constant y-velocity, then we have the following answer:

$v_{y} = -3.6\,\frac{m}{s}$

The y-velocity of the skateboard is -3.6 meters per second.