Final answer:
The particle travels a distance of B. 14 units when its velocity increases from 4 to 10 with a constant acceleration of 3.
Step-by-step explanation:
You asked how far a particle travels when its velocity increases from 4 to 10 with a constant acceleration of 3. To solve this, we can use the kinematic equation s = vt - 0.5at2, but we need to find the time (t) first. Since we know the initial velocity (u = 4), final velocity (v = 10), and acceleration (a = 3), we can use the formula v = u + at to find the time.
Given that the initial velocity (v_i) is 4, the final velocity (v_f) is 10, and the acceleration (a) is 3, we can substitute these values into the equation:
d = (10^2 - 4^2) / (2*3)
d = (100 - 16) / 6
d = 84 / 6
d = 14
Therefore, the particle travels a distance of 14 units.So, 10 = 4 + 3t, which gives us t = 2 seconds. We can then substitute this back into the displacement formula to get s = 4(2) + 0.5(3)(22) which equals 14 units. Therefore, the correct answer is (b) 14 units.
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