An object accelerates uniformly from rest and covers 8m in 2 seconds. What is the distance covered in the third second?

X(t)= (a/2) t^2

8 = (a/2)*2^2 = 2a
a = 4 m/s^2

After 3 seconds, the distance travelled from the starting point is
X(3) = 2*3^2 = 18 m
18 - 8 = 10 m was covered in the third second.
First second: deltaX = X(1) = 2 m
Second second: deltaX = X(2)-X(1) = 6 m
Third second: deltaX= X(3)-X(2) = 10 m

Well, it's a good thing this object is accelerating uniformly from rest, because if it was accelerating in a chaotic manner, things would get really messy! Anyway, let's solve this problem step by step, just like how a clown would solve a math problem.

In the first second, the object covers a distance of 8m divided by 2 seconds, which is, um... (counting on fingers) 4m. But don't worry, the math won't get too complicated, I promise!

Now, the object isn't going to stop after two seconds, because it's got some more distance to cover. So, for the third second, we can assume that the object will keep accelerating at the same rate.

To find the distance covered in the third second, we need to figure out how much the object accelerated in the first two seconds. Since the object is said to be accelerating uniformly, we can use a simple formula: distance equals half the acceleration times time squared. But wait, the object started from rest, so its initial velocity is zero! So the formula becomes distance equals half the acceleration times time squared.

Since the object covered 8m in 2 seconds, we can plug in the values and solve for acceleration: 8m equals half the acceleration times 2 seconds squared. Cracking out our handy-dandy calculator... (fake typing sounds)... we find that the acceleration is 4 meters per second squared.

Now, to find the distance covered in the third second, we use the formula again. This time, we calculate the distance covered in 3 seconds, subtracting the distance covered in the first 2 seconds.

So, distance in the third second equals half the acceleration times 3 seconds squared, minus 4 meters.

Calculating... (more fake typing)... drumroll, please... the distance covered in the third second is 18 meters!

So, the object covers 18 meters in the third second. And remember, math can be fun, just like a circus!

To find the distance covered in the third second, we first need to find the acceleration of the object.

Given:
Initial velocity (u) = 0 (object starts from rest)
Time (t) = 2 seconds
Distance covered (s) = 8 m

Using the formula:
s = ut + (1/2)at^2

Since the object starts from rest, the initial velocity (u) is 0, the equation simplifies to:
s = (1/2)at^2

We can rearrange the equation to isolate the acceleration (a):
a = (2s)/(t^2)

Substituting the given values:
a = (2 * 8)/(2^2)
a = 16/4
a = 4 m/s^2

Now, we can use this acceleration to find the distance covered in the third second.

Since the object starts from rest, its velocity at the end of the second second would be:
v = u + at
v = 0 + (4 m/s^2) * 2 s
v = 8 m/s

In the third second, the object's velocity will remain constant at 8 m/s.

Therefore, the distance covered in the third second can be calculated as:

distance = velocity * time
distance = 8 m/s * 1 s
distance = 8 m

Hence, the distance covered in the third second is 8 meters.

To find the distance covered in the third second, we first need to determine the average velocity of the object during the first two seconds.

We know that the object starts from rest, so its initial velocity (u) is 0 m/s.

The formula for average velocity is: average velocity (v) = (final velocity + initial velocity) / 2

Since the object is accelerating uniformly, we can calculate the final velocity (v) using the formula: final velocity (v) = (acceleration × time) + initial velocity.

The object covers 8 meters in 2 seconds, so the final velocity (v) can be calculated as follows:

8m = (acceleration × 2s) + 0

Simplifying the equation, we find that acceleration = 8m / 2s = 4m/s².

Now that we have the acceleration, we can calculate the final velocity (v):

v = (4m/s² × 2s) + 0
v = 8m/s.

The average velocity (v_avg) during the first two seconds is:

v_avg = (0 + 8m/s) / 2
v_avg = 4m/s.

So, in the first two seconds, the object had an average velocity of 4 m/s.

Now, to calculate the distance covered in the third second, we need to calculate the average velocity for that third second.

Since the acceleration is uniform, the velocity at the end of the second second is the same as the velocity at the beginning of the third second. Therefore, the velocity at the start of the third second is 8 m/s.

We can now calculate the distance covered in the third second using the equation: distance = (average velocity × time).

In this case, the average velocity is 8 m/s (since the velocity is constant during the third second), and the time is 1 second (since we want to find the distance covered in the third second):

distance = 8 m/s × 1 s
distance = 8 m.

Therefore, the object will cover a distance of 8 meters during the third second.