Enterprising students set an enormous slip-n-slide (a plastic sheet covered in water to reduce friction) on flat ground. If the slip-n-slide is 250 meters long. a. How small does the average acceleration have to be for a student starting at 5 meters/second to slide to the end?
b. If the acceleration is 0.34 m/s^2, what is the minimum initial speed a student would have to run to make it to the end? Is this speed possible?

Explanation:

a) Using the equation of motion

v² = u²+2as

v is the final velocity = 0m/s

u is the initial velocity = 5m/s

S is the distance = 250m

a is the acceleration

Substitute given values:

0² = 5²+2a(250)

-25 = 500a

a = -25/500

a = -0.05m/s²

b) If a = 0.34m/s²

We need to get the initial speed u

v² = u²+2as

0 = u²+2(0.34)(250)

-u² = 170

u² = -170

u = √-170

The square root of a negative number will give a complex number, hence the speed is not possible

  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.