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

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.

Step-by-step explanation:

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

`x(t) = x_(o) + v_(o,x)\cdot t + (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\,(m)/(s)`,
`t = 0.60\,s` and
`a_(x) = 1.8\,(m)/(s^(2))`, then the x-position of the skateboarder is:

`x(t) = 0\,m + \left(0\,(m)/(s) \right)\cdot (0.60\,s) + (1)/(2)\cdot \left(1.8\,(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 + (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\,(m)/(s)`,
`t = 0.60\,s` and
`a_(y) = 0\,(m)/(s^(2))`, then the x-position of the skateboarder is:

`y(t) = 0\,m + \left(-3.6\,(m)/(s) \right)\cdot (0.60\,s) + (1)/(2)\cdot \left(0\,(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\,(m)/(s)`,
`t = 0.60\,s` and
`a_(x) = 1.8\,(m)/(s^(2))`, then the x-velocity of the skateboarder is:

`v_(x)(t) = \left(0\,(m)/(s) \right) + \left(1.8\,(m)/(s) \right)\cdot (0.60\,s)`

`v_(x)(t) = 1.08\,(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\,(m)/(s)`

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