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Stolb23 [73]
2 years ago
5

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?
Physics
1 answer:
Dimas [21]2 years ago
3 0

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.

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