Question

A fish swimming in a horizontal plane has velocity v with arrowi = (4.00 î + 1.00 ĵ) m/s at a point in the ocean where the position relative to a certain rock is r with arrowi = (16.0 î − 3.00 ĵ) m. After the fish swims with constant acceleration for 15.0 s, its velocity is v with arrow = (25.0 î − 3.00 ĵ) m/s. (a) What are the components of the acceleration of the fish? ax = m/s2 ay = m/s2 (b) What is the direction of its acceleration with respect to unit vector î? ° counterclockwise from the +x-axis (c) If the fish maintains constant acceleration, where is it at t = 30.0 s? x = m y = m In what direction is it moving? ° counterclockwise from the +x-axis

Answer 1

The components of acceleration are ax = 21.0/15.0 m/s² and ay = -4.00/15.0 m/s². The direction of the acceleration with respect to the unit vector î is approximately -11.30° counterclockwise from the +x-axis. The fish is at the position x = 9586.0 m and y = -1773.0 m at t = 30.0 s and is moving approximately -6.77° counterclockwise from the +x-axis.

Step-by-step explanation:

(a) To find the components of acceleration, we can use the equation:

a = (v_f - v_i) / t

where a is acceleration, v_f is the final velocity, v_i is the initial velocity, and t is the time.

Using the given values:

v_i = (4.00 î + 1.00 ĵ) m/s

v_f = (25.0 î − 3.00 ĵ) m/s

t = 15.0 s

Plugging in the values, we get:

a = ((25.0 î − 3.00 ĵ) - (4.00 î + 1.00 ĵ)) / 15.0

= (21.0 î − 4.00 ĵ) / 15.0

= (21.0/15.0) î - (4.00/15.0) ĵ

So, the components of acceleration are:

ax = 21.0/15.0 m/s²

ay = -4.00/15.0 m/s²

(b) To find the direction of the acceleration with respect to the unit vector î, we can use the equation:

θ = tan^(-1)(ay/ax)

Plugging in the previously calculated values, we get:

θ = tan^(-1)(-4.00/21.0) ≈ -11.30°

So, the direction of the acceleration with respect to unit vector î is approximately -11.30° counterclockwise from the +x-axis.

(c) To find the position of the fish at t = 30.0 s, we can use the equation:

r = r_i + v_i*t + 0.5*a*t^2

where r is the position, r_i is the initial position, v_i is the initial velocity, a is the acceleration, and t is the time.

Using the given values:

r_i = (16.0 î − 3.00 ĵ) m

v_i = (4.00 î + 1.00 ĵ) m/s

t = 30.0 s

a = (21.0 î − 4.00 ĵ) m/s²

Plugging in the values, we get:

r = (16.0 î − 3.00 ĵ) + (4.00 î + 1.00 ĵ)*30.0 + 0.5*(21.0 î − 4.00 ĵ)*30.0^2

= (16.0 î − 3.00 ĵ) + (4.00 î + 1.00 ĵ)*30.0 + 0.5*(21.0 î − 4.00 ĵ)*900.0

= (16.0 î − 3.00 ĵ) + (120.0 î + 30.0 ĵ) + (9450.0 î − 1800.0 ĵ)

= (9586.0 î − 1773.0 ĵ)

So, the fish is at position:

x = 9586.0 m

y = -1773.0 m

Since the components of position are positive and negative, the fish is at the 2nd quadrant.

In what direction is it moving?

To find the direction of motion, we can use the equation:

θ = tan^(-1)(v_y/v_x)

Plugging in the previously calculated values, we get:

θ = tan^(-1)(-3.00/25.0) ≈ -6.77°

So, the fish is moving approximately -6.77° counterclockwise from the +x-axis.

Answer 2

a) ax = 1.40 m/s² ay = -0.267 m/s²

b) The direction of the acceleration vector is 349° with respect to the vector î measured counterclockwise from the +x-axis.

c) x = 776 m y = -93.2 m

The direction of the position vector at t = 30 s is 353° with respect to the vector î measured counterclockwise from the +x-axis.

Step-by-step explanation:

Please, see the attached figure 1 for a graphical description of the problem.

Since the acceleration is constant, the rate of change of the velocity can be written as follows:

dv/dt = a

Where :

dv/dt = variation of the velocity over time

a = acceleration.

Then:

dv = a dt

Integrating from v0 (initial velocity) to v (final velocity) and from t = 0 and t:

v - v0 = a · t

v = v0 + a · t

Then, using this equation we can obtain the components of the acceleration vector since we know the time and the components of the velocity vector:

vx = v0x + ax · t

25.0 m/s = 4.00 m/s + ax · 15.0 s

(25.0 m/s - 4.00 m/s) / 15.0 s = ax

ax = 1.40 m/s²

vy = v0y + ay · t

-3.00 = 1.00 m/s + ay · 15.0 s

(-3.00 m/s - 1.00 m/s) / 15.0 s = ay

ay = -0.267 m/s²

b) To find the direction of the vector acceleration, we have to find the angle θ in the figure. Then, the direction with respect to the unit vector î counterclockwise from the positive x-axis will be 270° + θ. Remember that the vector î = (1, 0).

Then, seeing the figure and using trigonometry:

cos θ = ay / a or sin θ = ax / a

Then, we have to calculate the magintude of the acceleration:

a² = ax² + ay² = (1.40 m/s²)² + (-0.267 m/s²)²

a = √((1.40 m/s²)² + (-0.267 m/s²)²)

a = 1.43 m/s²

Then (to avoid rounding error I will use the acceleration value without rounding):

sin θ = 1,40 m/s²/ √((1.40 m/s²)² + (-0.267 m/s²)²)

θ = 79.2°

Then, the direction of the acceleration vector will be:

270° + 79.2° = 349° with respect to the vector î measured counterclockwise from the +x-axis.

c)The procedure is similar to the points a) and b) only that now we have to use the equation of position instead of the equation of velocity. This equation is obtained by solving the following differential equation:

dx/dt = v = v0 + a · t

The equation of position is:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

a = accelertion

t = time

Then the components of the position vector (x,y) can be calcualted as follows:

x = x0 + v0x · t + 1/2 · ax · t²

y = y0 + v0y · t + 1/2 · ay · t²

Then:

x = 16.0 m + 4 m/s · 30 s + 1/2 · 1.40 m/s² · (30 s)²

x = 766 m

y = -3.00 + 1 m/s · 30 s - 1/2 · 0.267 m/s² · (30 s)²

y = -93.2 m

At t = 30 s, the fish is at x = 766 m and y = -93.2 m relative to the rock.

To find the direction we need the magnitude of the vector r:

The magnitude of the vector position will be:

r² = x² + y² = (766 m)² + (-93.2 m)²

r = 772 m

Please see the figure 2 to notice that sin θ = x / r and that the direction of r will be 270° + θ

Then:

sin θ = 766 m/ √( (766 m)² + (-93.2 m)²)

θ = 83.1°

The direction of the position vector will be (270° + 83.1°) 353° with respect to the vector î measured counterclockwise from the +x-axis.