Solution:

Given, v = 5 - t (m/s)

To find the distance covered by the particle in the first 10 seconds, we need to integrate the velocity function with respect to time from 0 to 10 seconds.

∫v dt = ∫(5-t) dt

= 5t - (t^2/2) + C

where C is the constant of integration.

At t = 0, the particle is at rest, so the constant of integration C is zero.

Therefore, the distance covered in the first 10 seconds is

∫(0 to 10) v dt = [5t - (t^2/2)](0 to 10)

= [5(10) - (10^2/2)] - [5(0) - (0^2/2)]

= (50 - 50) - (0 - 0)

= 0 m

Oops! There seems to be some mistake in the given answer. Let's check our calculation.

The distance covered by the particle is the magnitude of the displacement vector, which is given by the area under the velocity-time graph.

To find the area under the graph, we can divide the time interval into small intervals of Δt and approximate the area of each small rectangle as vΔt.

Then, the total area can be approximated as the sum of the areas of all the rectangles.

Let's take Δt = 1 second.

At t = 0, v = 5 m/s. So, the area of the first rectangle is

vΔt = (5 m/s)(1 s) = 5 m

At t = 1, v = 4 m/s. So, the area of the second rectangle is

vΔt = (4 m/s)(1 s) = 4 m

Similarly, we can find the areas of the remaining rectangles.

At t = 9, v = -4 m/s. So, the area of the tenth rectangle is

vΔt = (-4 m/s)(1 s) = -4 m

At t = 10, v = -5 m/s. So, the area of the last rectangle is

vΔt = (-5 m/s)(1 s) = -5 m

The negative sign indicates that the particle is moving in the opposite direction.

Therefore, the total area under the graph is

5 + 4 + 3 + 2 + 1 + 0 + (-1) + (-2) + (-3) + (-4) + (-5)

= 0 m

Oops! There seems to be some mistake in our approximation. Let's try a better method.

We can approximate the area under the graph as a trapezium with height equal to the average velocity and base equal to the time interval.

The average velocity in the first second is (5 + 4)/2 = 4.5 m/s.

The average velocity in the second second is (4 + 3)/2 = 3.5 m/s.

Similarly, we can find the average velocities in the remaining seconds.

The area of the trapezium is

(1/2)(4.5 + 3.5)(1 s) + (1/2)(3.5 + 2.5)(1 s) + ... + (1