A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet.
f(t) = te−t/4
(a) Find the velocity at time t (in ft/s).
v(t) =
(b) What is the velocity after 1 s? (Round your answer to two decimal places.)
v(1) = ft/s
(c) When is the particle at rest?
t = s
(d) When is the particle moving in the positive direction? (Enter your answer using interval notation.)
(e) Find the total distance traveled during the first 10 s. (Round your answer to two decimal places.)
ft
(f) Find the acceleration at time t (in ft/s2).
a(t) =
Find the acceleration after 1 s. (Round your answer to three decimal places.)
a(1) = ft/s2
(g) Graph the position, velocity, and acceleration functions for 0 ≤ t ≤ 10.
To find the velocity at time t, we need to take the derivative of the position function with respect to time:
(a) v(t) = f'(t)
The derivative of f(t) = te^(-t/4) can be found using the chain rule. The chain rule states that if we have a function g(x) = f(u(x)), then the derivative of g(x) with respect to x is g'(x) = f'(u(x)) * u'(x).
Applying the chain rule, we have:
f'(t) = (1) * e^(-t/4) + t * (-1/4) * e^(-t/4)
Simplifying, we get:
f'(t) = e^(-t/4) - t/4 * e^(-t/4)
This is the velocity function at time t.
(b) To find the velocity after 1 second, we substitute t = 1 into the velocity function:
v(1) = e^(-1/4) - (1/4) * e^(-1/4)
Evaluate this expression to find the velocity in feet per second.
(c) The particle is at rest when its velocity is 0. So, we need to solve the equation v(t) = 0 for t:
0 = e^(-t/4) - t/4 * e^(-t/4)
To solve this equation, we can make use of numerical methods such as Newton's method or use a calculator or software capable of solving equations numerically.
(d) The particle is moving in the positive direction when its velocity is positive. In other words, when v(t) > 0. Determine the intervals in which v(t) > 0 to find when the particle is moving in the positive direction.
(e) To find the total distance traveled during the first 10 seconds, we need to integrate the absolute value of the velocity function over the interval [0,10]:
Total distance = ∫[0,10] |v(t)| dt
Integrate the absolute value of v(t) over the interval [0,10] using calculus techniques or a calculator or software capable of calculating definite integrals numerically.
(f) To find the acceleration at time t, we need to take the derivative of the velocity function with respect to time:
a(t) = v'(t)
Differentiate the velocity function we obtained in part (a) to find the acceleration function.
(g) To graph the position, velocity, and acceleration functions for 0 ≤ t ≤ 10, plot the respective functions on a graph with time (t) on the x-axis and the corresponding value (s, v, or a) on the y-axis. You can use graphing software or an online graphing tool to plot the functions.
(a) To find the velocity at time t, we need to take the derivative of the position function f(t):
f(t) = te^(-t/4)
Using the product rule of differentiation, we have:
f'(t) = 1 * e^(-t/4) + t * (-1/4) * e^(-t/4)
= e^(-t/4) - (t/4) * e^(-t/4)
Therefore, the velocity function v(t) is:
v(t) = e^(-t/4) - (t/4) * e^(-t/4)
(b) To find the velocity after 1 second, we can substitute t = 1 into the velocity function:
v(1) = e^(-1/4) - (1/4) * e^(-1/4)
Using a calculator, we can find the approximate value of v(1) as:
v(1) ≈ -0.276 ft/s
(c) The particle is at rest when its velocity is zero. So, we can set v(t) = 0 and solve for t:
e^(-t/4) - (t/4) * e^(-t/4) = 0
Factorizing out e^(-t/4), we get:
e^(-t/4) * (1 - t/4) = 0
Either e^(-t/4) = 0 (which has no solution) or (1 - t/4) = 0.
Solving (1 - t/4) = 0 for t, we have:
1 - t/4 = 0
t/4 = 1
t = 4
Therefore, the particle is at rest at t = 4 seconds.
(d) The particle is moving in the positive direction when its velocity is positive. From the velocity function v(t) = e^(-t/4) - (t/4) * e^(-t/4), we can see that the particle is moving in the positive direction when e^(-t/4) > (t/4) * e^(-t/4).
Simplifying this inequality, we get:
1 > t/4
4 > t
So, the particle is moving in the positive direction for values of t less than 4 (t < 4) in interval notation.
(e) To find the total distance traveled during the first 10 seconds, we can integrate the absolute value of the velocity function over the interval [0, 10]:
∫[0,10] |v(t)| dt
This gives us the definite integral of |v(t)| over the interval from 0 to 10. Evaluating this integral will give us the desired distance.
(f) To find the acceleration at time t, we need to take the derivative of the velocity function v(t):
v(t) = e^(-t/4) - (t/4) * e^(-t/4)
Using the product rule of differentiation, we have:
v'(t) = (-1/4) * e^(-t/4) - (t/4) * (-1/4) * e^(-t/4) + (1/4) * e^(-t/4)
= (-1/4) * e^(-t/4) + (t/16) * e^(-t/4) + (1/4) * e^(-t/4)
= (-1/4 + t/16 + 1/4) * e^(-t/4)
= (t/16) * e^(-t/4)
Therefore, the acceleration function a(t) is:
a(t) = (t/16) * e^(-t/4)
(g) To graph the position, velocity, and acceleration functions for 0 ≤ t ≤ 10, we will need to plot the corresponding values for each function over this interval. However, without specific values for t, we cannot provide a complete graph.