A speed skater moving across frictionless ice at 8.8 m/s hits a 6.0 m -wide patch of rough ice. She slows steadily, then continues on at 5.8 m/s . What is her acceleration on the rough ice?

Recall that

v² - u² = 2 a ∆x

where u and v are initial and final velocities, respectively; a is acceleration; and ∆x is the distance traveled (because the skater moves in only one direction).

So we have

(5.8 m/s)² - (8.8 m/s)² = 2 a (6.0 m)

a = ((5.8 m/s)² - (8.8 m/s)²) / (12 m)

a = -3.65 m/s²

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.