An ant moves along the x-axis with velocity given by v(t)=tsin(t^2), t is greater than or equal to zero.
Given that x(t) is the position of the particle at time t and that x(0)=3, find x(2)
Find the total distance traveled by the particle from t=0 to t=2
This is an initial value problem, where you would solve a differential equation (or an integral), and with a given set of initial values, determine the integration constant.
Let x(t)=position at time t
v(t)=tsin(t^2)
∫v(t)dt = ∫tsin(t²)dt
Integral of velocity gives distance
For the right-hand side, substitute u=t^2,
du=2tdt
x(t)=∫(1/2)sin(u)du
=-cos(u)/2+C'
=-cos(t^2)+C
Given x(0)=3
=>
x(0)=-cos(0)+C = 3
C=3+cos(0)=4
=>
x(t)=-cos(t^2)+4
x(2)=4-cos(2^2)=4-cos(4)
Distance travaelled = x(2)-x(0)=?
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To find x(t), we need to integrate the velocity function v(t) with respect to t.
∫v(t) dt = x(t) + C (C is the constant of integration)
Given v(t) = tsin(t^2), we can integrate it:
∫(tsin(t^2)) dt
To integrate this, we can use a u-substitution, where u = t^2.
Taking the derivative of both sides with respect to t, we get du/dt = 2t, which implies dt = (1/2t)du.
Substituting these values back into the integral:
∫(tsin(t^2)) dt = ∫sin(u) * (1/2t) du = (1/2)∫sin(u)/t du
Using the fact that ∫sin(u) du = -cos(u), we have:
(1/2)∫sin(u)/t du = (-1/2)cos(u) + C
Substituting back u = t^2:
(1/2)∫sin(u)/t du = (-1/2)cos(t^2) + C
Therefore, the position function x(t) is:
x(t) = (-1/2)cos(t^2) + C
Given that x(0) = 3, we can determine the constant of integration C:
x(0) = (-1/2)cos(0^2) + C
3 = (-1/2)(1) + C
3 = -1/2 + C
C = 3 + 1/2
C = 7/2
Thus, the position function x(t) is:
x(t) = (-1/2)cos(t^2) + 7/2
To find x(2), we substitute t = 2 into the equation:
x(2) = (-1/2)cos((2)^2) + 7/2
x(2) = (-1/2)cos(4) + 7/2
Calculating this expression will give us the value of x(2).
To find the total distance traveled by the particle from t = 0 to t = 2, we need to calculate the definite integral of the absolute value of the velocity function ∣v(t)∣ over the interval [0, 2]:
∫[0,2]∣v(t)∣ dt
Since the velocity function is positive for t > 0, the absolute value is unnecessary. Therefore, we have:
∫[0,2] v(t) dt
To evaluate this definite integral, we can simply use the antiderivative of v(t) and evaluate it over the interval [0,2]:
∫[0,2] v(t) dt = [x(t)] from 0 to 2
Substituting the position function x(t), we have:
∫[0,2] v(t) dt = [(-1/2)cos(t^2) + 7/2] from 0 to 2
Evaluating this expression will give us the total distance traveled by the particle.