A ball is pushed down a hill with an initial velocity of 2.5 m/s. As it rolls it accelerates at a rate of 6m/s/s. What will be it’s velocity 1.5 minutes later?

Step-by-step explanation:

This is more physics than math because you need to know the equations for one-dimensional motion to get this done. The equation we need is

where v is the final velocity, v0 is the initial velocity, a is the acceleration and t is the time in seconds. That means that since our time is given in minutes we have to convert it to seconds. I'm not sure how much importance your teacher gives to significant digits, but if you use them here, the answer you get is way off, so we will disregard them here. 1.5 minutes is 90 seconds. Now that have the time in seconds and the equation we need, we will fill it in accordingly:

Note immediately that the seconds on the time cancels out one of the seconds in the denominator of the acceleration so we're left with m/sec + m/sec which is what we need to add those 2 terms. Multiply 6 by 90 to get 540 m/sec. then add that to 2.5 m/sec to get 542.5 m/sec. That must be really long hill!!!

y = 3x+4

step-by-step explanation:

to find the slope

slope = (y2-y1)/ (x2-x1)

    = (-8-4)/ (-4-0)

    = -12/-4

  = 3

using the slope intercept form of a line

y= mx+b

we know the slope is 3

we know the y intercept is 4 ( the point where it crosses the y axis or x=0)

y = 3x+4