Two-Dimensional Motion with Constant Acceleration Problems and Solutions

 Problem#1

At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of v_i = (3.00i - 2.00j) m/s and is at the origin. At t = 3.00 s, the particle’s velocity is v_f = (9.00i + 7.00j) m/s. Find (a) the acceleration of the particle and (b) its coordinates at any time t.

Answer:
(a) the acceleration of the particle is
v_f = v_i + at

(9.00i + 7.00j) m/s = (3.00i - 2.00j) m/s + a(3.00 s)

a = (2.00i + 3.00j) m/s²

(b) particle position is given by
r_f = r_i + v_it + ½ at²

r_f = (3.00i - 2.00j)t + ½ (2.00i + 3.00j)t²

r_f = [(3.00t + t²)i + (1.5t² – 2.00t)j] m

then, its coordinates at any time t, is

x = (3.00t + t²) m and y = (1.5t² – 2.00t) m

Problem#2
The vector position of a particle varies in time according to the expression r = (3.00i - 6.00t² j) m.
(a) Find expressions for the velocity and acceleration as functions of time.
(b) Determine the particle’s position and velocity at t = 1.00 s.

Answer:
(a) expressions for the velocity and acceleration as functions of time given by

For velocity:
v = dr/dt = d[(3.00i - 6.00t² j)]/dt = –12.0tj m/s

for acceleration:

a = dv/dt = d[–12.0tj]/dt = –12.0j m/s²

(b) the particle’s position at t = 1.00 s is

r = 3.00i – 6.00(1.00)²j = (3.00i – 6.00j) m

and the particle’s velocity at t = 1.00 s is

v = –12.0j m/s

Problem#3
A fish swimming in a horizontal plane has velocity v_i = (4.00i + 1.00j) m/s at a point in the ocean where the position relative to a certain rock is r_i = (10.0i - 4.00j) m. After the fish swims with constant acceleration for 20.0 s, its velocity is v = (20.0i - 5.00j) m/s.
(a) What are the components of the acceleration?
(b) What is the direction of the acceleration with respect to unit vector i?
(c) If the fish maintains constant acceleration, where is it at t = 25.0 s, and in what direction is it moving?

Answer:
Given v_i = (4.00i + 1.00j) m/s and v (t = 20 s) = (20.0i - 5.00j) m/s

(a) the components of the acceleration given by

a_x = ∆v_x/∆t = {(20.0i - 5.00j) - (4.00i + 1.00j)}/20 s

a_x = (0.800i – 0.300j) m/s²

then a_x = 0.800 m/s² and a_y = -0.300 m/s²

(b) the direction of the acceleration with respect to unit vector I, given by

θ = tan^-1(a_y/a_x) = tan^-1[-0.300/0.800] = -20.6⁰ or 330⁰ from +x axis

(c) at t = 25.0 s

x_f = x_i + v_ixt + ½ a_xt² = 10.0 + 4.00(25) + ½ (0.800)(25)² = 360 m

y_f = y_i + v_iyt + ½ a_yt² = -4.00 + 1.00(25) + ½ (-0.300)(25)² = -72.7 m
and
v_xf = v_xi + a_xt = 4 + 0.08 x 25 = 24 m/s

v_yf = v_yi + a_yt = 1 - 0.03 x 25 = -6.5 m/s

θ = tan^-1(v_y/v_x) = tan^-1[-6.5/24] = -15.2⁰ or 344.8⁰ from +x axis

Problem#4
A particle initially located at the origin has an acceleration of a = 3.00j m/s² and an initial velocity of v_i = 500i m/s. Find :
(a) the vector position and velocity at any time t and
(b) the coordinates and speed of the particle at t = 2.00 s.

Answer:
Given: a = 3.00j m/s², v_i = 5.00i m/s

(a) the vector position at any time t given by

r_f = r_i + v_it + ½ at²

r_f = 0 + (5.00it + ½ (3.00j)t²

r_f = (5.00ti + 1.50j) m

and velocity at any time t given by

v_f = v_i + at

v_f = (5.00i) m/s + (3.00j)t

v_f = (5.00i + 3.00tj) m/s

(b) the coordinates of the particle at t = 2.00 s.

At t = 2.00 s

r_f = (5.00 x 2.00 i + 1.50j) m = (10.0i + 1.5j)

then (x,y) = (10.0, 1.5) m

and speed of the particle at t = 2.00 s.

v_f = (5.00i + 3.00tj) m/s

v_f = (5.00i + 3.00 x 2.00j) m/s

v_f = (5.00i + 6.00j) m/s

then,

v_f = √(v_x² + v_y²)

v_f = √(5.00² + 6.00²) = 7.81 m/s

Problem#5
It is not possible to see very small objects, such as viruses, using an ordinary light microscope. An electron microscope can view such objects using an electron beam instead of a light beam. Electron microscopy has proved invaluable for investigations of viruses, cell membranes and subcellular structures, bacterial surfaces, visual receptors, chloroplasts, and the contractile properties of muscles. The “lenses” of an electron microscope consist of electric and magnetic fields that control the electron beam. As an example of the manipulation of an electron beam, consider an electron traveling away from the origin along the x axis in the xy plane with initial velocity v_i = v_ii . As it passes through the region x = 0 to x = d, the electron experiences acceleration a = a_xi + a_yj, where ax and ay are constants. For the case v_i = 1.80 x 10⁷ m/s, a_x = 8.00 x 10¹⁴ m/s² and a_y = 1.60 x 10¹⁵ m/s², determine at x = d = 0.0100 m
(a) the position of the electron,
(b) the velocity of the electron,
(c) the speed of the electron, and
(d) the direction of travel of the electron (i.e., the angle between its velocity and the x axis).

Answer:
(a) the position of the electron,

For the x-component of the motion we have
x_f = x_i + v_xit + ½ a_xt²

0 = 0 + (1.80 x 10⁷ m/s)t + ½ (8 x 10¹⁴)t²

or

(4 x 10¹⁴)t² + (1.80 x 10⁷)t – 10^-2 = 0

 

We choose the + sign to represent the physical situation
t = (4.39 x 10⁵ m/s)/(8 x 10¹⁴ m/s²) = 5.49 x 10^-10 s

Here
y_f = y_i + v_yit + ½ a_yt² = 0 + 0 + ½ (1.6 x 10¹⁵ m/s²)(5.49 x 10^-10 s)² = 2.41 x 10^-4 m
so,
r = xi + yj = (10.0i + 0.241j) mm

(b) the velocity of the electron,

v_f = v_i + at

v_f = (1.80 x 10⁷ m/s)i + (8 x 10¹⁴ m/s²i + 1.6 x 10¹⁵ m/s²j)(5.49 x 10^-10 s)

v_f = (1.84 x 10⁷ m/s)i + (8.78 x 10⁵ m/s)j

(c) the speed of the electron, and

v_f = √(v_x² + v_y²)

v_f = √[(1.84 x 10⁷)² + (8.78 x 10⁵)²] = 1.84 x 10⁷ m/s

(d) the direction of travel of the electron (i.e., the angle between its velocity and the x axis).

θ = tan^-1(v_y/v_x) = tan^-1[(8.78 x 10⁵)/(1.84 x 10⁷)] = -2.73⁰  from +x axis