Finding equation, displacement, distance

Question

The acceleration of an object (in m/s²) is given by the function a(t)=5sin(t). The initial velocity of the object is v(0)=−1m/s. Round your answers to four decimal places.

a) Find an equation v(t) for the object velocity.

v(t)=   

b) Find the object's displacement (in meters) from time 0 to time 3.

   meters

c) Find the total distance traveled by the object from time 0 to time 3.

   meters

Answer 1

this problem requires the use of indefinite integrals and fundamental theorem of calculus (FTC) part 2 to solve.

a) given the problem is asking for an equation of velocity v(t), and the initial information provided is the equation of acceleration a(t), we must use indefinite integrals to solve this problem.

note that with higher derivatives, the notation of each derivative is as follows using f(t):

f(t) = position

f’(t) = velocity

f’’(t) = acceleration

this means that the acceleration equation is your second derivative function and the velocity equation will be your first derivative function. note that d/dx -cos x = sin x. because 5 serves as a coefficient to the function, it will not be affected by finding derivatives or antiderivatives.

f’’(t) = a(t) = 5sin t

f’(t) = v(t) = ∫(5sin t) dt

 = -5cos t + C

usually, C can be any real number, but in this case, this information is given to us to find what C equals. using the knowledge that v(0) = -1 m/s, plug this into the function found above to find the velocity equation to solve for C:

-1 = -5cos(0) + C

-1 = -5 + C

C = 4

thus, v(t) = -5cos t + 4

b) using the steps above, we must now find the position function s(t).

f(t) = s(t) = ∫₀³](-5cos t + 4) dt

= -5sin t + 4t

now, plug in the endpoints [0, 3] into the equation using FTC part 2:

displacement = [-5sin(3) + 4(3)] – [-5sin(0) + 4(0)] = [-5sin(3) + 12] – 0 = 11.2945 meters

c) because velocity becomes positive after starting negative, we must find where v(t) = 0:

-5cos t + 4 = 0 —> cos t = 0.8 —> t₀ = cos^-1(0.8) = 0.6435

thus, velocity is negative on [0, 0.6435) and positive on (0.6435, 3].

now, we must find distance traveled on both intervals using FTC part 2, then add the answers together:

s₁ = ∫₀^t0(-5cos t + 4) dt = -5sin t + 4t  evaluated at [0, 0.6435]

s₁ = [-5sin(0.6435) + 4(0.6435)] – [-5sin(0) + 4(0)] = -0.4260

distance cannot be negative so: | s1 | = | -0.4260 | = 0.4260 meters

let total displacement = 11.2945 meters

s₂ = 11.2945 – (-0.4260) = 11.7205

thus, total distance = s₁ + s₂ = 11.7205 + 0.4260 = 12.1465 meters