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2 years ago

8

A skateboarder travels on a horizontal surface with an initial velocity of 3.2 m/s toward the south and a constant acceleration

of 2.4 m/s2 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?

1 answer:

Naddika [18.5K]2 years ago

8 0

Answer:

A. 0.432

B. -1.92

C. 1.44 units/second

D. -3.2 units/second

Explanation:

A. To calculate her x position, we just use the following equation of motion to find the distance traveled:

$s=u*t+\frac{1}{2} (a*t^2)$

here s = displacement

t = time (in seconds)

a = acceleration

Solving for the distance, we get:

$s = 0 * 0.6 + \frac{1}{2}(2.4 * 0.6^2)$

s = 0.432 m

Since 0.432 meters east is equals to 0.432 meter in the positive x-direction, the x position is also 0.432.

B. Since the skater has a constant v - velocity of -3.2 m/s, (south means negative y axis), the total distance traveled is:

Distance = speed * time = -3.2 * 0.6 = -1.92 m

The answer is -1.92 units in the y-axis.

C. The x velocity component is the final speed in the east direction, which is going to be:

$v^2 - u^2=2*a*s$

$v^2 = 2*2.4*0.432$

v = 1.44 units/second (in positive x direction)

D. Her y velocity component does not change, since the velocity towards the south is a constant 3.2 m/s

Thus the answer is -3.2 units/second in the y-axis.

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