5. A particle moves along the y – axis with velocity given by v(t)=tsine(t^2) for t>=0 .
a. In which direction (up or down) is the particle moving at time t = 1.5? Why?
b. Find the acceleration of the particle at time t= 1.5. Is the velocity of the particle increasing at t = 1.5? Why or why not?
c. Given that y (t) is the position of the particle at time t and that y (0) = 3, find y (2).
d. Find the total distance traveled by the particle from t = 0 to t = 2.
v(t) = t sin t^2
a(t) = t( cos t^2) (2t) + sin t^2
= 2t^2 cos t^2 + sin t^2
y(t) = (1/2) (-cos t^2) + c
given : when t = 0 , y(0) = 3
3 = (-1/2)cos 0 + c
3 = -1/2 + c
c = 3.5
so y(t) = (-1/2) cos t^2 + 3.5
You now have enough to answer all the other questions.
Remember that if the first derivative is posititive, the function is increasing, (if a(t) > 0, then v(t) increases)
y(2) = (-1/2) cos (4) + 3.5 = appr 4.8
y(0) = 3
so the distance traveled from t-0 to t-2 = 4.8-3 = 1.8
To answer these questions, we need to analyze the given velocity function and use basic calculus concepts. Here's how we can solve each part:
a. To determine the direction (up or down) of the particle's motion at t = 1.5, we need to find the sign of the velocity v(t) at that time. Substitute t = 1.5 into the velocity equation:
v(1.5) = 1.5sin(1.5^2)
You can use a calculator to evaluate this expression. The result will give you the value of the velocity at t = 1.5. If the velocity is positive, the particle is moving up; if the velocity is negative, the particle is moving down.
b. The acceleration of the particle at a given time t can be found by differentiating the velocity function with respect to t:
a(t) = d(v(t))/dt
Differentiate v(t) = tsin(t^2) using the chain rule:
a(t) = sin(t^2) + 2t^2cos(t^2)
Evaluate a(1.5) by substituting t = 1.5 into the acceleration equation. This will give you the acceleration of the particle at t = 1.5.
To determine if the velocity of the particle is increasing at t = 1.5, check the sign of the acceleration. If the acceleration is positive, the velocity is increasing; if it's negative, the velocity is decreasing.
c. To find y(2), we need to integrate the velocity function with respect to t to obtain the position function, y(t):
y(t) = ∫(v(t))dt
Integrate v(t) = tsin(t^2) using substitution:
y(t) = -0.5cos(t^2) + C
Given that y(0) = 3, substitute t = 0 and solve for C:
3 = -0.5cos(0) + C
C = 3.5
Now, plug in t = 2 into the position function:
y(2) = -0.5cos(2^2) + 3.5
Evaluate this expression using a calculator to find the position of the particle at t = 2.
d. 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 over the interval [0, 2]:
distance = ∫(|v(t)|)dt from 0 to 2
This represents the area under the velocity graph between t = 0 and t = 2. Calculate this integral using the given velocity function and the appropriate limits of integration.
Once you evaluate this integral, you will obtain the total distance traveled by the particle.
a. To determine the direction of the particle's movement at time t = 1.5, we need to evaluate the velocity at that time.
Plugging in t = 1.5 into the velocity function, we have:
v(1.5) = 1.5 * sin((1.5)^2)
To determine if the particle is moving up or down, we need to consider the sign of v(1.5). If v(1.5) is positive, then the particle is moving up. If v(1.5) is negative, then the particle is moving down.
b. To find the acceleration of the particle at time t = 1.5, we need to take the derivative of the velocity function with respect to time.
a(t) = d/dt(v(t))
= d/dt(t * sin(t^2))
Differentiating the function, we get:
a(t) = sin(t^2) + 2t^2cos(t^2)
To determine if the velocity of the particle is increasing at t = 1.5, we need to evaluate the acceleration at that time. If a(1.5) is positive, then the velocity is increasing. If a(1.5) is negative, then the velocity is decreasing.
c. To find the position of the particle at time t = 2, we need to integrate the velocity function with respect to time.
y(t) = ∫ v(t) dt + C
The constant C can be found by using the initial condition y(0) = 3.
d. To find the total distance traveled by the particle from t = 0 to t = 2, we need to consider both the positive and negative portions of the velocity function. The total distance traveled can be found by taking the integral of the absolute value of the velocity function from t = 0 to t = 2:
distance = ∫ |v(t)| dt from 0 to 2.