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 ]
The answer is 2.5 times heavier than on Earth !!
so the answer is C !!
Answer: yea ma’am I’m sorry but you still
Answer:
-589.05 J
Explanation:
Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner
So, W = ΔK
W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)
So, substituting the values of the variables into the equation, we have
W = 1/2m(v₁² - v₀²)
W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)
W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)
W = 1/2 × 72.9 kg(-16.1604 m²/s²)
W = 1/2 × (-1178.09316 kgm²/s²)
W = -589.04658 kgm²/s²
W = -589.047 J
W ≅ -589.05 J
m = mass = 1,200 kg
A = acceleration = 3 m/s^2
Apply Newton's second law:
Force = mass x acceleration
F = 1,200 x 3 =3600 N
The net force the car experiences is 3600 N