Velocity Of An Object: \( V = 9t^2 + 4t + 6 \)

The Velocity Of An Object Is Given By Vt 9t2 4t 6 At T 1 S T

The velocity of an object is given by v(t) = 9t² - 4t + 6. At t = 1 s, the position of the object is at 10 meters. In this context, the tasks are to determine the acceleration equation and the position equation for the object, derive the velocity-position relationship v² = v₀² + 2aΔx, analyze motion scenarios involving deceleration and minimum acceleration requirements, and examine a specific object with a given position equation to find specific time points and positions.

Paper For Above instruction

This paper aims to analyze and solve multiple problems related to kinematic equations, including deriving the acceleration and position equations for an object with a specified velocity function, establishing the fundamental velocity-position relation, and applying these concepts to real-world scenarios such as a bicycle deceleration, an aircraft's takeoff dynamics, and a train’s stopping distance. Finally, it involves analyzing the motion of a given object with a specified position equation to find velocities at specific times and positions when acceleration is zero.

Deriving the Acceleration and Position Equations

Given the velocity function v(t) = 9t² - 4t + 6, the acceleration a(t) can be directly obtained by differentiating v(t) with respect to time t:



a(t) = dv/dt = d/dt [9t² - 4t + 6] = 18t - 4

Hence, the acceleration equation is a(t) = 18t - 4.

To find the position equation x(t), integrate v(t):



x(t) = ∫ v(t) dt = ∫ (9t² - 4t + 6) dt = 3t³ - 2t² + 6t + C

where C is the constant of integration, determined by initial conditions. Given that at t = 1 s, x = 10 m:



x(1) = 3(1)³ - 2(1)² + 6(1) + C = 3 - 2 + 6 + C = 7 + C

Setting x(1) = 10 m:



7 + C = 10 => C = 3

Therefore, the position equation is:



x(t) = 3t³ - 2t² + 6t + 3

Deriving the Velocity-Position Equation v² = v₀² + 2aΔx

The equation connecting velocity and displacement arises from the basic kinematic equations. Starting with the acceleration definition a = dv/dt and noting that v = dx/dt, we can write:



a = dv/dt = (dv/dx)(dx/dt) = v(dv/dx)

Rearranged:



v dv = a dx

Integrating both sides:



∫ v dv = ∫ a dx

Assuming a constant acceleration a, the integrals are:



(1/2) v² = a x + C

Applying initial conditions at v = v₀ when x = x₀, the constant C becomes (1/2) v₀² - a x₀. Therefore, the general equation simplifies to:



v² = v₀² + 2a (x - x₀)

This is the well-known velocity-position relation used extensively in kinematics.

Deceleration of a Bicycle

A cyclist on a recumbent bicycle slows down from a constant velocity of 40 m/s over 5 seconds. The deceleration (negative acceleration) can be calculated as:



a = Δv / Δt = (0 - 40) / 5 = -8 m/s²

The negative sign indicates deceleration.

The distance traveled before coming to a stop is found using:



Δx = v₀ t + (1/2) a t² = 40 * 5 + (1/2)(-8)(5)^2 = 200 - 100 = 100 meters

Jet Takeoff Dynamics

A Cessna jet (90-RC) requires a minimum acceleration to lift off within 240 meters at a takeoff velocity of 120 km/hr (which is 33.33 m/s). The minimum acceleration is calculated by rearranging the equation:



v² = 2 a x

Solving for a:



a = v² / (2 x) = (33.33)^2 / (2 * 240) ≈ 1111.11 / 480 ≈ 2.31 m/s²

The time to reach takeoff velocity at this acceleration:



t = v / a = 33.33 / 2.31 ≈ 14.44 seconds

Train Stopping Distance and Time

A train traveling at 100 m/s applies brakes with an acceleration of -2.2 m/s². The time to stop:



t = (final velocity - initial velocity) / acceleration = (0 - 100) / -2.2 ≈ 45.45 seconds

The distance covered during braking:



Δx = v₀ t + (1/2) a t² = 100 * 45.45 + (1/2)(-2.2)(45.45)^2 ≈ 4545 - 2250 ≈ 2295 meters

Motion of an Object with Position Equation x(t) = (2/3)t^3 - 4t² + 6t + 4

(a) To find when the velocity is zero, first find the velocity function by differentiating position:



v(t) = dx/dt = 2 t^2 - 8 t + 6

Set v(t) = 0:



2 t^2 - 8 t + 6 = 0 => t^2 - 4 t + 3 = 0

Solve quadratic:



t = [4 ± √(16 - 12)] / 2 = [4 ± 2] / 2

Hence, t = (4 + 2)/2 = 3 seconds and t = (4 - 2)/2 = 1 second.

(b) To find positions where acceleration is zero, first find acceleration:



a(t) = d v(t)/dt = d/dt [2 t^2 - 8 t + 6] = 4 t - 8

Set a(t) = 0:



4 t - 8 = 0 => t = 2 seconds

Evaluate position x(t) at t = 2 seconds:



x(2) = (2/3)(8) - 4(4) + 6(2) + 4 = (16/3) - 16 + 12 + 4 ≈ 5.33 - 16 + 12 + 4 ≈ 5.33

Thus, at t = 2 seconds, the acceleration is zero, and the position is approximately 5.33 meters.

Conclusion

This analysis illustrates fundamental principles of kinematics through deriving equations and applying them to various physical situations. The derived acceleration and position functions enable precise understanding of motion dynamics, while the problem-specific analyses demonstrate the practical use of these equations in everyday and engineering contexts. The ability to derive and manipulate kinematic equations is central to solving a broad array of physics problems and understanding the motion of objects in real-world scenarios.

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