Having trouble with this questions. Please help.
A particle moves on the x-axis so that its position at any time t (is greater than or equal to) 0 is given by x(t) = 2te^-t
a) Find the acceleration of the particle at t=0
b)find the velocity of the particle when its acceleration is 0.
c) find the total distance traveled by the particle from t=0 to t=5
Damon is wrong. The acceleration equation would be 2te^-t - 4e^-t.
v = dx/dt = -2 t (e^-t) + 2 e^-t
a = d^2x/dt^2 = 2 t e^-t + 2 e^-t - 2 e^-t = 2 t e^-t
so
a) at t = 0, a = 2(0) e^0 = 0
b) v(0) = 0 + 2 (1) = 2
c) Does the particle turn around and head back between t = 0 and t = 5?
v = 0 = -2 t e^-t +2 e^-t
t = 1 when the sign of v reverses
when t = 1, x = 2 e^-1 = .736
so from t = 0 to t = 1 it went .736 positive
now from t = 1 to t = 5
x(5) = 10 e^-10
= tiny
so it came back essentially to zero
so total distance = .736 forward + .736 in reverse = 1.47
Sure, I'd be happy to help you with your questions! Let's tackle them one by one:
a) To find the acceleration of the particle at t = 0, we need to find the second derivative of the position function x(t).
So, let's start by finding the velocity of the particle:
v(t) = (dx(t)/dt)
Taking the first derivative of x(t), we get:
x'(t) = 2(e^(-t) - t(e^(-t)))
Now, let's find the acceleration of the particle:
a(t) = (dv(t)/dt)
Taking the derivative of v(t), we get:
v'(t) = -2e^(-t) + te^(-t)
Now, we can find the acceleration at t = 0 by substituting t = 0 into a(t):
a(0) = -2e^(0) + 0e^(0)
= -2 + 0
= -2
Therefore, the acceleration of the particle at t = 0 is -2.
b) To find the velocity of the particle when its acceleration is 0, we need to find the values of t where a(t) = 0.
Setting a(t) to 0 and solving for t:
-2e^(-t) + te^(-t) = 0
To solve this equation, we can factor out e^(-t):
e^(-t)(-2 + t) = 0
Setting each factor to 0:
e^(-t) = 0 (This is not possible, as e^(-t) is always positive)
-2 + t = 0
Solving the second equation for t, we find:
t = 2
Therefore, when the acceleration is 0, the velocity of the particle is obtained at t = 2.
c) To find the total distance traveled by the particle from t = 0 to t = 5, we need to calculate the definite integral of the absolute value of the velocity function over the interval [0, 5].
Using the velocity function we previously found:
v(t) = -2e^(-t) + te^(-t)
The total distance traveled is given by:
Total distance = ∫[0,5]|v(t)| dt
To evaluate the integral, we need to consider two cases:
- When t < 2: |v(t)| = -(v(t)) since v(t) < 0
- When t > 2: |v(t)| = v(t) since v(t) > 0
Let's calculate the integral for each case:
For t < 2:
∫[0,2] -(v(t)) dt = -∫[0,2] (-2e^(-t) + te^(-t)) dt
For t > 2:
∫[2,5] v(t) dt = ∫[2,5] (-2e^(-t) + te^(-t)) dt
Evaluating these integrals will give you the total distance traveled by the particle from t = 0 to t = 5.
Remember, integration can sometimes be hairy, just like a wild clown wig. So take your time and don't forget to embrace the laughter along the way!
Sure! I can help you with these questions. Let's break them down one by one.
a) To find the acceleration of the particle at t=0, we need to find the second derivative of the position function x(t). The second derivative gives us the rate of change of acceleration.
To start, let's find the first derivative of x(t):
x'(t) = d/dt (2te^-t)
Using the product rule of differentiation:
x'(t) = 2(e^-t)(1) + (-2te^-t)(1) = 2e^-t - 2te^-t
Now, let's find the second derivative by taking the derivative of x'(t):
x''(t) = d/dt (2e^-t - 2te^-t)
Using the product rule once again, we get:
x''(t) = (-2e^-t) + (2te^-t)(1) + (-2e^-t)(1) = -4e^-t + 2te^-t
Now, substitute t=0 into x''(t) to find the acceleration at t=0:
x''(0) = -4e^0 + 2(0)e^0 = -4 + 0 = -4
Therefore, the acceleration of the particle at t=0 is -4.
b) To find the velocity of the particle when its acceleration is 0, we need to find the value of t for which x''(t) = 0.
Using the equation we found for the second derivative:
-4e^-t + 2te^-t = 0
Factor out e^-t:
e^-t (-4 + 2t) = 0
Since e^-t is always positive, we can set -4 + 2t = 0:
2t = 4
t = 2
So, when the acceleration is 0, the velocity of the particle is given by x'(2). Let's find it:
x'(2) = 2e^-2 - 2(2)e^-2
x'(2) = 2e^-2 - 4e^-2 = -2e^-2
Therefore, when the acceleration is 0, the velocity of the particle is -2e^-2.
c) To find the total distance traveled by the particle from t=0 to t=5, we need to find the integral of the absolute value of the velocity function over the interval [0, 5].
The absolute value of the velocity function is |x'(t)| = |-2e^-t| = 2e^-t.
Now, let's find the integral:
∫(0 to 5) 2e^-t dt
To integrate this, we can use the antiderivative of e^-t, which is -e^-t. Thus:
∫(0 to 5) 2e^-t dt = 2(-e^-t) ∣ (0 to 5) = 2(-e^-5 + e^-0) = 2(1 - 1/e^5)
Therefore, the total distance traveled by the particle from t=0 to t=5 is 2(1 - 1/e^5) units.