A rabbit runs in a garden such that the x- and y-components of its displacement as function of times are given by
x(t) = −0.44t2 − 6.0t + 73 and y(t)= 0.35t2 + 8.3t + 34.
(Both x and y are in meters and t is in seconds.)
(a) Calculate the rabbit's position (magnitude and direction) at t = 9 s.
magnitude 1
direction 2° north of west
(b) Calculate the rabbit's velocity at t = 9 s.
magnitude 3
direction 4° north of west
(c) Determine the acceleration vector at t = 9 s.
aarrowitalic = m/s2
a) 174.1 m
b) 21.8 m/s
c) 114 m/s2
(a) Well, if we plug in t = 9 s into the equations x(t) and y(t), we can find the rabbit's position. Let's do the math:
x(9) = -0.44 * (9)^2 - 6.0 * 9 + 73
= -0.44 * 81 - 54 + 73
= -35.64 - 54 + 73
= -16.64
y(9) = 0.35 * (9)^2 + 8.3 * 9 + 34
= 0.35 * 81 + 74.7 + 34
= 28.35 + 74.7 + 34
= 137.05
So, the rabbit's position at t = 9 s is actually (-16.64, 137.05) meters.
To calculate the magnitude of the position, we can use the Pythagorean theorem:
magnitude = sqrt((-16.64)^2 + (137.05)^2)
= sqrt(275.9296 + 18775.7025)
= sqrt(19051.6321)
≈ 138.01 meters
Now, let's calculate the direction. We'll use trigonometry:
direction = arctan(137.05 / -16.64)
≈ arctan(-8.23359)
≈ -2° east of south
Oops, it seems the rabbit took a wrong turn and went east instead of west. I hope it didn't get lost!
(b) To find the rabbit's velocity at t = 9 s, we need to differentiate the x(t) and y(t) equations with respect to time (t).
vx(t) = -0.44 * 2t - 6.0
= -0.88t - 6.0
vy(t) = 0.35 * 2t + 8.3
= 0.70t + 8.3
Now, let's substitute t = 9 s:
vx(9) = -0.88 * 9 - 6.0
= -7.92 - 6.0
= -13.92
vy(9) = 0.70 * 9 + 8.3
= 6.3 + 8.3
= 14.6
The rabbit's velocity at t = 9 s is (-13.92, 14.6) m/s.
To calculate the magnitude of the velocity, we again use the Pythagorean theorem:
magnitude = sqrt((-13.92)^2 + (14.6)^2)
= sqrt(193.7664 + 213.16)
≈ sqrt(406.9264)
≈ 20.17 m/s
Now, let's calculate the direction:
direction = arctan(14.6 / -13.92)
≈ arctan(-1.04943)
≈ -45.8° east of south
Well, it seems the rabbit is still going east instead of west. Someone needs to give it a map!
(c) To determine the acceleration vector at t = 9 s, we differentiate the velocity equations with respect to time (t).
ax(t) = -0.88
ay(t) = 0.70
The rabbit's acceleration vector at t = 9 s is simply (-0.88, 0.70) m/s^2. No need for further calculations here!
So, the acceleration vector at t = 9 s is approximately (-0.88, 0.70) m/s^2. The rabbit may be accelerating, but I'm sure he's hopping to his own rhythm!
To solve this problem, we'll need to find the position, velocity, and acceleration of the rabbit at t = 9 s. Let's calculate each part step by step.
(a) Position at t = 9 s
To find the position of the rabbit, we need to substitute t = 9 into the x(t) and y(t) equations.
x(9) = -0.44(9)^2 - 6(9) + 73
= -0.44(81) - 54 + 73
= -35.64 - 54 + 73
= -16.64 meters
y(9) = 0.35(9)^2 + 8.3(9) + 34
= 0.35(81) + 74.7 + 34
= 28.35 + 74.7 + 34
= 137.05 meters
Therefore, the position of the rabbit at t = 9 s is (-16.64, 137.05) meters.
To find the magnitude of the position, we use the Pythagorean theorem:
Magnitude = sqrt((-16.64)^2 + (137.05)^2)
= sqrt(276.8896 + 18767.3025)
= sqrt(19044.1921)
= 137.994 meters
To find the direction, we can use the inverse tangent function:
Direction = arctan(137.05 / -16.64)
≈ -68.01°
Therefore, the rabbit's position at t = 9 s is approximately 137.994 meters, 68.01° south of east.
(b) Velocity at t = 9 s
To find the velocity at t = 9 s, we need to differentiate the x(t) and y(t) equations with respect to time (t).
vx(t) = d/dt (-0.44t^2 - 6.0t + 73)
= -0.88t - 6.0
vy(t) = d/dt (0.35t^2 + 8.3t + 34)
= 0.70t + 8.3
Now, let's calculate the velocity at t = 9 s:
vx(9) = -0.88(9) - 6.0
= -7.92 - 6.0
= -13.92 m/s
vy(9) = 0.7(9) + 8.3
= 6.3 + 8.3
= 14.6 m/s
Therefore, the velocity of the rabbit at t = 9 s is approximately (-13.92, 14.6) m/s.
To find the magnitude of the velocity, we use the Pythagorean theorem:
Magnitude = sqrt((-13.92)^2 + (14.6)^2)
= sqrt(193.7664 + 213.16)
= sqrt(406.9264)
= 20.174 m/s
To find the direction, we can use the inverse tangent function:
Direction = arctan(14.6 / -13.92)
≈ -44°
Therefore, the rabbit's velocity at t = 9 s is approximately 20.174 m/s, 44° south of west.
(c) Acceleration at t = 9 s
To find the acceleration at t = 9 s, we need to differentiate the velocity equations with respect to time (t).
ax(t) = d/dt (-0.88t - 6.0)
= -0.88
ay(t) = d/dt (0.70t + 8.3)
= 0.70
Therefore, the acceleration of the rabbit at t = 9 s is (-0.88, 0.70) m/s^2.
To solve this problem, we need to find the position, velocity, and acceleration of the rabbit at t = 9s.
(a) To find the position of the rabbit at t = 9s, we need to find the x and y coordinates separately and then combine them.
At t = 9s:
x(t) = -0.44(9)^2 - 6.0(9) + 73
x(t) = -0.44(81) - 54 + 73
x(t) = -35.64
y(t) = 0.35(9)^2 + 8.3(9) + 34
y(t) = 0.35(81) + 74.7 + 34
y(t) = 28.35 + 74.7 + 34
y(t) = 137.05
The position vector of the rabbit at t = 9s is (x(t), y(t)) = (-35.64, 137.05).
To find the magnitude and direction of the position vector, we can use the Pythagorean theorem and trigonometry.
Magnitude: magnitude = sqrt((-35.64)^2 + (137.05)^2)
Magnitude: magnitude = sqrt(1273.6896 + 18795.6025)
Magnitude: magnitude = sqrt(20069.2921)
Magnitude: magnitude ≈ 141.67
Direction: To find the direction, we can find the angle using arctan(dy/dx).
Angle: angle = arctan((137.05)/(-35.64))
Angle: angle ≈ 104.7°
Therefore, the rabbit's position at t = 9s is approximately 141.67 meters at an angle of 104.7°.
(b) To find the velocity of the rabbit at t = 9s, we need to find the derivatives of x(t) and y(t) with respect to time.
At t = 9s:
v_x(t) = -0.44(2)(9) - 6.0
v_x(t) = -7.92
v_y(t) = 0.35(2)(9) + 8.3
v_y(t) = 6.3 + 8.3
v_y(t) = 14.6
The velocity vector of the rabbit at t = 9s is (v_x(t), v_y(t)) = (-7.92, 14.6).
To find the magnitude and direction of the velocity vector:
Magnitude: magnitude = sqrt((-7.92)^2 + (14.6)^2)
Magnitude: magnitude = sqrt(62.7264 + 213.16)
Magnitude: magnitude ≈ 15.80
Direction: Angle: angle = arctan((14.6)/(-7.92))
Angle: angle ≈ 118.5°
Therefore, the rabbit's velocity at t = 9s is approximately 15.80 m/s at an angle of 118.5°.
(c) To find the acceleration vector at t = 9s, we need to find the second derivatives of x(t) and y(t) with respect to time.
At t = 9s:
a_x(t) = -0.44(2) = -0.88
a_y(t) = 0.35(2) = 0.7
The acceleration vector of the rabbit at t = 9s is (a_x(t), a_y(t)) = (-0.88, 0.7) m/s².