At the moment t=0 a particle starts moving along thr x-axis so that its velocity projection varies as vx= 35 cos pi t cm/s, where 't' is expressed in seconds. Find the distance that this particle covers during t = 2.80 s after the start.
It will move back and forth along the x axis with frequency 0.5 Hz.
If distance covered is defined as
DeltaX = xfinal - xinitial, then
Delta X = Integral vx dt
0 to 2.8 s
= 35 pi sin(2.8 pi)
The particle will be in its second back-and-forth period, and will have "covered" 4 amplitudes during the first period.
To find the distance that the particle covers during t = 2.80 s after the start, we need to integrate the velocity function with respect to time from t = 0 to t = 2.80.
The given velocity projection function is vx = 35 cos(pi t) cm/s.
To integrate this function, we follow these steps:
1. Apply the power rule of integration: ∫(cos(x)) dx = sin(x) + C, where C is the constant of integration.
2. Apply the chain rule to account for the πt within the cosine function: ∫(cos(πt)) dt = (∫cos(πt)) * (dt/dt) = (∫cos(πt)) * 1
3. Integrate the function: ∫(35 cos(πt)) dt = (35/π) sin(πt) + C
The indefinite integral is (35/π) sin(πt) + C. Now we can find the definite integral by evaluating the antiderivative at the upper and lower limits of integration.
To find the distance covered during t = 2.80 s after the start, we evaluate the definite integral from t = 0 to t = 2.80:
Distance covered = ∫[0 to 2.80] (35/π) sin(πt) dt
Substituting the limits into the antiderivative, we have:
Distance covered = [(35/π) sin(π * 2.80)] - [(35/π) sin(π * 0)]
Simplifying further:
Distance covered = (35/π) sin(2.80π) - 0
Using a calculator to evaluate sin(2.80π), we get:
Distance covered = (35/π) * sin(8.806) cm
Approximating this value, the distance covered during t = 2.80 s after the start is approximately 24.423 cm.