A force of 1.50 N acts on a 0.20kg trolley so as to accelerate it along an air track The track and force are horizontal and in line . How fast is the trolley going after acceleration from rest through 30cm , if friction is negligible

2.12m/s

Explanation:

Given parameters:

Force on trolley  = 1.5N

Mass of trolley  = 0.2kg

Unknown:

Velocity of the trolley  = ?

Solution:

To solve this problem, we first find the acceleration of the trolley;

          Force  = mass x acceleration

         Acceleration  =

Insert the parameters and solve;

          Acceleration  =   = 7.5m/s²

Now to find the acceleration;

  Initial velocity  = 0m/s

  v² = u² + 2aS

v is the final velocity

u is the initial velocity

a is the acceleration

S is the distance

  Distance  = 30cm and this is 0.3m

     v² = 0² + 2(7.5)0.3  = 4.5

     v = √4.5 = 2.12m/s

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14