An object is travelling with a constant acceleration of 10 ms⁻² . how much distance will it travel in 3rd second of its journey?

Given:

Acceleration is uniform and acceleration (a) = 10 m/s^2

Now it has been mentioned for the first 2 secs the acceleration is 10m/s^2 .

Hence velocity= acceleration x time

Velocity= 10 x 2 = 20 m/s

Consider s as the distance traveled in the 3rd second.

Now we know s= ut+1/2(at^2)

Where s is the distance measured in m.

u is the initial velocity measured in m/sec

t is the time taken for the object to travel the above distance. This is equal to one second as we need to calculate the distance traveled between 3rd and 2nd second.

t = (3-2)= 1 sec

Substituting the given values in the above formula we get

s = 20 x1 + 1/2 (10 x 1 x 1)

s = 25 m

Thus the distance traveled by the object in the 3rd second is 25 m

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

c.the charges if the proton and electron and gravity