A particle moves along a line so that its velocity at time t is ( | |. Find the displacement of the particle during the time period .

Complete question is;

A particle moves along a line so that its velocity at time t is

v(t) = t² − t − 6

(measured in meters per second).

(a) Find the displacement of the particle during the time period 1 ⩽ t ⩽ 4.

(b) Find the distance traveled during this time period.

A) -4.5 m

B) 10.17 m

Explanation:

We are given;

v(t) = t² − t − 6

We know that v = ds/dt

Thus,

S = v dt

s(4) - s(1) = (4,1)∫t² − t − 6 dt

= (4,1)[(t³/3) - (t²/2) - 6t]

= ((4³/3) - (4²/2) - 6(4)) - ((1³/3) - (1²/2) - 6(1))

= 64/3 - 8 - 24 - 1/3 + 1/2 + 6

= -4.5 m

B) Since v(t) = t² − t − 6, then factorizing we can write it as;

v(t) = (t + 2)(t - 3)

Thus, v(t) ⩽ 0 at the interval (1, 3) and v(t) ≥ 0 at the interval (3, 4)

Thus;

-x1 = (1, 3)∫t² − t − 6 dt

-x1 = (1,3)(t³/3 - t²/2 - 6t)

-x1 = (3³/3 - 3²/2 - 6(3)) - (1³/3 - 1²/2 - 6(1))

-x1 = 9 - 9/2 - 18 - 1/3 + 1/2 + 6

-x1 = -3 - 4 - 1/3

-x1 = -22/3 m

x1 = 22/3 m

x2 = (3, 4)∫t² − t − 6 dt

x2 = (3, 4)(t³/3 - t²/2 - 6t)

x2 = ((4³/3) - (4²/2) - 6(4)) - (3³/3 - 3²/2 - 6(3))

x2 = 64/3 - 8 - 24 - 9 + 9/2 + 18

x2 = -23 + 64/3 + 9/2

x2 = (-138 + 128 + 27)/6

x2 = 17/6 m

Thus, total distance = x1 + x2 = 22/3 + 17/6 = 10.17m

the order for strong to weak forces such as

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force

the fundamental forces are those forces that act between to objects or element to make interactions. there are four types of fundamental forces such as strong and weak nuclear forces (due to protons and electrons), electromagnetic force (due to the motion of electrons) and gravitational force (due to orbital movement of the planet). if these forces are arranged on the basis of their maximum strength then the order is

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force

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