9. A particle moves in the x-y plane with a y-component of velocity in meter per second given
by Vy =8t with t in seconds. The acceleration of the particle in the x-direction in meter per
second squared is given by ax =4twith t in seconds. When t=0, y=2m, x=0, and Vx=0. Find
the equation of the path of the particle and calculate the magnitude of the velocity v of the
particle for the instant when its x coordinate reaches 8*041m.
10.A particle moves in the X –Y plane so that its x coordinate is defined by
x = 5 t^3 – 105 t where x is in cm and t is in seconds. When t = 2 s the total acceleration is
7*041m/s2. If the y component of acceleration is constant and the particles starts from rest at
the origin when t = 0, determine its velocity when t = 4 sec.
11.A particle moving in the x-y plane has a position vector given by where
r is in Cm and t is in seconds. Calculate the radius of curvature of the path for the position
of the particle when t=0*041sec.
9. To find the equation of the path of the particle, we first need to find the x-component of velocity and acceleration. Since Vx = dx/dt, we can integrate ax = 4t with respect to t to find Vx:
Vx = ∫(4t)dt
Vx = 2t^2 + C
Given that Vx = 0 when t = 0, we can solve for C:
0 = 2(0)^2 + C
C = 0
So, Vx = 2t^2
Now, to find the x-component of position, we integrate Vx with respect to t:
x = ∫(2t^2)dt
x = (2/3)t^3 + D
Given that x = 0 when t = 0, we can solve for D:
0 = (2/3)(0)^3 + D
D = 0
So, the x-component of position is x = (2/3)t^3
Now, we can find the y-component of position as y = ∫(8t)dt:
y = 4t^2 + E
Given that y = 2 when t = 0, we can solve for E:
2 = 4(0)^2 + E
E = 2
So, the y-component of position is y = 4t^2 + 2
Therefore, the equation of the path of the particle is x = (2/3)t^3, y = 4t^2 + 2.
To calculate the magnitude of the velocity v when x = 8*041m, we can differentiate the equations for x and y with respect to time to find Vx and Vy, respectively:
Vx = dx/dt = 2t^2
Vy = dy/dt = 8t
Now, we can calculate the magnitude of the velocity v using the formula v = √(Vx^2 + Vy^2):
v = √((2t^2)^2 + (8t)^2)
v = √(4t^4 + 64t^2)
Given that x = 8*041m, we can solve for t:
(2/3)t^3 = 8*041
t = 2
Substitute t = 2 into the equation for v:
v = √(4(2)^4 + 64(2)^2)
v = √(4(16) + 64(4))
v = √(64 + 256)
v = √320
v ≈ 17.89 m/s
So, the magnitude of the velocity of the particle is approximately 17.89 m/s when its x coordinate reaches 8*041m.
11. In order to calculate the radius of curvature of the path for the position of the particle when t = 0.041 sec, we need to first find the velocity and acceleration vectors of the particle.
Given the position vector r(t) = (5t + 3) i + (4t - t^2) j, we can find the velocity vector V(t) by differentiating the position vector with respect to time:
V(t) = (d/dt)(5t + 3) i + (d/dt)(4t - t^2) j
V(t) = 5i + (4 - 2t) j
Next, we differentiate the velocity vector to find the acceleration vector A(t):
A(t) = (d/dt)(5)i + (d/dt)(4 - 2t) j
A(t) = 0i - 2j
When t = 0.041 sec, the acceleration vector A(0.041) = -2j. The magnitude of acceleration is the magnitude of the acceleration vector, which is given by |A| = √((-2)^2) = 2 m/s^2.
The radius of curvature (R) can be calculated using the formula:
R = |V|^3 / |A|
Given that |V| = √(5^2 + (4 - 2(0.041))^2) = √(25 + 3.766) = √28.766 ≈ 5.36 m/s,
we can now calculate the radius of curvature:
R = (5.36)^3 / 2 ≈ 76.98 m
Therefore, the radius of curvature of the path for the position of the particle when t = 0.041 sec is approximately 76.98 meters.