physics CH3+CH4

Question

A fish swimming in a horizontal plane has velocity _i = (4.00 î + 1.00 ĵ) m/s at a point in the ocean where the position relative to a certain rock is _i = (14.0 î − 2.80 ĵ) m. After the fish swims with constant acceleration for 19.0 s, its velocity is  = (23.0 î − 1.00 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 = 26.0 s?

x =

m

y =

m

In what direction is it moving?

 ° counterclockwise from the +x-axis

Answer 1

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

Δv = aΔt

where Δv is the change in velocity, a is the acceleration, and Δt is the time interval.

The change in velocity is:

Δv = v_f - v_i = (23.0 î - 1.00 ĵ) m/s - (4.00 î + 1.00 ĵ) m/s = (19.0 î - 2.00 ĵ) m/s

Plugging in the values, we get:

a = Δv / Δt = (19.0 î - 2.00 ĵ) m/s / 19.0 s = (1.00 î - 0.105 ĵ) m/s^2

Therefore, the components of the acceleration of the fish are:

ax = 1.00 m/s^2 ay = -0.105 m/s^2

(b) The direction of the acceleration with respect to the unit vector î can be found using the formula:

θ = tan^-1(ay / ax)

Plugging in the values, we get:

θ = tan^-1(-0.105 / 1.00) = -5.29 degrees

Therefore, the direction of the acceleration is 5.29 degrees counterclockwise from the +x-axis.

(c) To find the position of the fish at t = 26.0 s, we can use the equations:

Δx = v_i,t Δt + 1/2 a Δt^2 Δy = v_i,y Δt + 1/2 a Δt^2

where Δx and Δy are the changes in position, v_i,t is the initial velocity in the x-direction, v_i,y is the initial velocity in the y-direction, and a is the acceleration.

At t = 19.0 s, the position of the fish relative to the rock is:

r = (14.0 î - 2.80 ĵ) m

The initial velocity in the x-direction is:

v_i,t = 4.00 m/s

The initial velocity in the y-direction is:

v_i,y = 1.00 m/s

Plugging in the values, we get:

Δx = v_i,t Δt + 1/2 a Δt^2 = (4.00 m/s)(26.0 s - 19.0 s) + 1/2 (1.00 m/s^2) (26.0 s - 19.0 s)^2 = 98.0 m

Δy = v_i,y Δt + 1/2 a Δt^2 = (1.00 m/s)(26.0 s - 19.0 s) + 1/2 (-0.105 m/s^2) (26.0 s - 19.0 s)^2 = 2.37 m

Therefore, the fish is at a position of:

x = r_x + Δx = 14.0 m + 98.0 m = 112.0 m y = r_y + Δy = -2.80 m + 2.37 m = -0.43 m

The fish is moving in the direction of its velocity, which is given by:

θ = tan^-1(v_y / v_x) = tan^-1(-1.00 / 23.0) = -2.51 degrees

Therefore, the fish is moving in the direction of -2.51 degrees counterclockwise from the +x-axis.