A force of 1.50 N acts on a 0.20 kg 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 30 cm, if friction is negligible

F = ma

v^2 = 2as

To find the final velocity of the trolley, we can use the equation of motion:

v^2 = u^2 + 2as

Where:
v = final velocity
u = initial velocity (0 m/s since the trolley starts from rest)
a = acceleration
s = distance traveled

First, let's convert the distance from centimeters to meters:

s = 30 cm = 0.30 m

Next, we need to calculate the acceleration using Newton's second law:

F = ma

Rearranging the formula, we have:

a = F/m

Substituting the given values:

a = 1.50 N / 0.20 kg

a = 7.5 m/s^2

Now, we can calculate the final velocity:

v^2 = 0^2 + 2 * 7.5 * 0.30

v^2 = 0.90

v = √(0.90)

v ≈ 0.95 m/s

Therefore, the trolley is moving at approximately 0.95 m/s after accelerating from rest through 30 cm.