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

F = ma

v^2 = 2as

Good

To solve this problem, you can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.

Given:
Force (F) = 1.50 N
Mass (m) = 0.20 kg
Initial velocity (u) = 0 (as the cart is at rest)
Displacement (s) = 30 cm = 0.30 m
Friction (negligible)

Step 1: Find the acceleration (a) of the cart.
Use the formula F = ma and rearrange it to solve for a:
a = F / m

a = 1.50 N / 0.20 kg
a = 7.50 m/s^2

Step 2: Use the kinematic equation to find the final velocity (v) of the cart.
The kinematic equation that relates initial velocity (u), final velocity (v), acceleration (a), and displacement (s) is:
v^2 = u^2 + 2as

Plugging in the given values:
v^2 = 0^2 + 2(7.50 m/s^2)(0.30 m)
v^2 = 0 + 4.50 m^2/s^2
v = √(4.50 m^2/s^2)
v ≈ 2.12 m/s

Therefore, the cart is traveling at approximately 2.12 m/s after accelerating through a distance of 30 cm.