In the diagram, q1 = +6.60x10^-6 C, q2 = +3.10x10^-6 C, and q3 = +5.30x10^-6 C. Find the electric potential energy of q2. (Remember, energy is NOT a vector.)

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explanation:

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

b. a lightning bug, plug of grass, and a worm