it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]
Answer:
v = 4.76 m/s
Explanation:
Given,
The distance traveled by her bike, d = 10 miles
The time of her travel, t = 2.1 m/s
The velocity of an object is defined as the distance traveled by the object to the time of travel. Therefore,
V = d/t m/s
= 10 / 2.1
= 4.76 m/s
Hence, The velocity of her bike is, V = 4.76 m/s
Answer:
x sin nx = x cos nx
same as
theta / theta x (xsin (nx)) = sin (nx) + (nx) cos (nx)
Answer:
Correct answer: C. 50 cm
Explanation:
Given data:
The distance of the object from the top of the concave mirror o = 50.0 cm
The magnitude of the concave mirror focal length 25.0 cm.
Required : Image distance d = ?
If we know the focal length we can calculate the center of the curve of the mirror
r = 2 · f = 2 · 25 = 50 cm
If we know the theory of spherical mirrors and the construction of figures then we know that when an object is placed in the center of the curve, there is also a image in the center of the curve that is inverted, real and the same size as the object.
We conclude that the image distance is 50 cm.
We will now prove this using the formula:
1/f = 1/o + 1/d => 1/d = 1/f - 1/o = 1/25 - 1/50 = 2/50 - 1/50 = 1/50
1/d = 1/50 => d = 50 cm
God is with you!!!
Answer:
ω₂ = 1.9025 x 10⁻⁶ rad/s
Explanation:
given,
mass of star = 1.61 x 10³¹ kg
angular velocity = 1.60 x 10⁻⁷ rad/s
diameter suddenly shrinks = 0.29 x present size
r₂ = 0.29 r₁
using conservation of angular momentum
I₁ ω₁ = I₂ ω₂
ω₂ = 1.9025 x 10⁻⁶ rad/s
