Physics College

The following equation shows the position of a particle in time t, x=at2i + btj where t is in second and x is in meter. A=2m/s2, b=1m/s.
Find

A, the average velocity of the particle in the time interval t₁=2sec and t₂=3sec

B, the velocity and acceleration at any time t.

C, the average acceleration in the time interval given in part (a)​

Answers

Answer 1

(a) The average velocity of the particle in the time interval t₁=2sec and t₂=3sec is 10 m/s.

(b) The velocity and acceleration at any time t is v =  (4ti + j) m/s and a = a = 4i m/s²

(c)  The average acceleration in the time interval given in part (a)​ is 3.98 m/s².

Position of the particle

x = at²i + btj

x = 2t²i + tj

Average velocity, at t₁=2sec and t₂=3sec

Δv = Δx/Δt

x(2) = 2(2)²i + 2j

x(2) = 8i + 2j

|x(2)| = √(8² + 2²) = 8.246

x(3) =  2(3)²i + 3j

x(3) = 18i + 3j

|x(3)| = √(18² + 3²) = 18.248

Δv = (18.248 - 8.246)/(3 - 2)

Δv =  10 m/s

Velocity and acceleration at any time, t

v = dx/dt

v =  (4ti + j) m/s

a = dv/dt

a = 4i m/s²

Average acceleration

v(2) = 4(2)i + j

v(2) = 8i + j

|v(2)| = 8.06 m/s

v(3) = 4(3)i + j

v(3) = 12i + j

|v(3)| = 12.04 m/s

a = (12.04 - 8.06)/(3 - 2)

a = 3.98 m/s²

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Related Questions

The resistivity of pure water is fairly high, 1.8 × 105 Ω·m, whereas the resistivity of sea water is a million times lower, 2 × 10-1 Ω·m. Why does the high salt concentration make sea water significantly less resistive (i.e. more conductive) than pure water?

Answers

The high salt concentration make sea water significantly less resistive than pure water due to presence of charged ions in the sea water.

What is resistivity?

The resistivity of a substance is the opposition to the flow of charges offered by the substance.

The greater the resistivity of a substance, the lesser its conductivity.

A high salt concentration make sea water highly conductive due to presence of charged ions in the water. The greater conductivity reduces the resistivity of the sea water.

A pure water has no charged ions in the water, thereby decreasing its conductivity and increasing its resistivity.

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A 3,204 kg tree positioned on the edge of a cliff 247 m above the ground breaks away and falls into the valley below which is considered zero potential energy. If the tree’s mechanical energy is conserved, what is the speed of the tree just before it hits the ground in meters/sec?

Answers

Let's see

PE is turned to KE as per law of conservation of energy

[tex]\\ \rm\Rrightarrow mgh=\dfrac{1}{2}mv^2[/tex]

[tex]\\ \rm\Rrightarrow 2gh=v^2[/tex]

[tex]\\ \rm\Rrightarrow 2(10)(247)=v^2[/tex]

[tex]\\ \rm\Rrightarrow v²=4940[/tex]

[tex]\\ \rm\Rrightarrow v=70.3ms^{-1}[/tex]

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