Some Easy Ones!

From Eckerd Academic Wiki

1. A particle moves along the x-axis starting from rest at the origin. Four seconds later it is at x = 8 meters moving at a velocity of 10 m/s.

(a) Find the average velocity.

<v> = x/t = (x_f - x_i)/(t_f - t_i) = (8 - 0)/(4 - 0) = 2.0 m/s

(b) Find the average acceleration.

<a> = v/t = (v_f - v_i)/(t_f - t_i) = (10 - 0)/(4 - 0) = 2.5 m/s²

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2. A particle moves along the x-axis. At time t = 2 sec it is located at x = - 2.0 meters moving in the negative x direction at 4.0 m/s. Later, at t = 7 sec it is located at x = - 17.0 meters moving in the negative x direction at 10 m/s.

(a) Find the average velocity.

<v> = x/t = (x_f - x_i)/(t_f - t_i) = [-17.0 - (-2)]/(7 - 2) = - 3.0 m/s

(b) Find the average acceleration.

<a> = v/t = (v_f - v_i)/(t_f - t_i) = (-10.0 - (-4)]/(7 - 2) = - 1.2 m/s²

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3. A particle's position along the x-axis is given by x(t) = t³ - 2t² - 6t - 2, where x is in meters and t in seconds.

(a) Find the particle's position at t = 2 sec and at t = 4 sec.

x(2) = 2³ - (2)(2)² - (6)(2) - 2 = - 14.0 m

x(4) = 4³ - (2)(4)² - (6)(4) - 2 = + 6.0 m

(b) Find the particle's instantaneous velocity at t = 2 sec and at t = 4 sec, and its average velocity for the time interval (2,4).

v(t) = dx/dt = 3t² - 4t - 6

Therefore:

v(2) = (3)(2)² - (4)(2) - 6 = - 2.0 m/s

v(4) = (3)(4)² - (4)(4) - 6 = + 26.0 m/s

<v> = x/t = (x_f - x_i)/(t_f - t_i) = [6.0 - (-14)]/(4 - 2) = + 10 m/s

(c) Find the particle's instantaneous acceleration at t = 2 sec and at t = 4 sec, and its average acceleration for the time interval (2,4).

a(t) = dv/dt = 6t - 4

Therefore:

a(2) = (6)(2) - 4 = + 8.0 m/s²

a(4) = (6)(4) - 4 = + 20.0 m/s²

<a> = v/t = (v_f - v_i)/(t_f - t_i) = [26.0 - (- 2.0)]/(4 - 2) = + 14.0 m/s²

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