Find the position and velocity of an object moving along a straight line with the given acceleration, initial velocity,
and initial position.
, and

Ask by Brooks Smith. in the United States

Mar 20,2025

Question

Find the position and velocity of an object moving along a straight line with the given acceleration, initial velocity,
and initial position.
, and

Ask by Brooks Smith. in the United States

Mar 20,2025

UpStudy AI · Accepted Answer

To find the position and velocity of an object moving along a straight line with the given acceleration, initial velocity, and initial position, we can use the following equations:

1. Acceleration equation: $$ a(t) = \frac{dv}{dt} $$
2. Velocity equation: $$ v(t) = \frac{ds}{dt} $$
3. Position equation: $$ s(t) = \int v(t) dt $$

Given:
- Acceleration equation: $$ a(t) = 4e^{-t} $$
- Initial velocity: $$ v(0) = -4 $$
- Initial position: $$ s(0) = 8 $$

We can find the velocity equation by integrating the acceleration equation with respect to time:
$$ v(t) = \int a(t) dt = \int 4e^{-t} dt $$

Now, let's calculate the velocity equation.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int 4e^{-t} dt$$
- step1: Use properties of integrals:
  $$4\times \int e^{-t} dt$$
- step2: Evaluate the integral:
  $$4\left(-e^{-t}\right)$$
- step3: Multiply the terms:
  $$-4e^{-t}$$
- step4: Add the constant of integral C:
  $$-4e^{-t} + C, C \in \mathbb{R}$$
The velocity equation is $$ v(t) = -4e^{-t} + C $$, where $$ C $$ is a constant of integration.

Given that the initial velocity is $$ v(0) = -4 $$, we can substitute $$ t = 0 $$ into the velocity equation to find the value of $$ C $$:
$$ v(0) = -4e^{-0} + C = -4 + C = -4 $$
$$ C = 0 $$

Therefore, the velocity equation is $$ v(t) = -4e^{-t} $$.

Now, let's find the position equation by integrating the velocity equation with respect to time:
$$ s(t) = \int v(t) dt = \int -4e^{-t} dt $$

Next, we will calculate the position equation.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int -4e^{-t} dt$$
- step1: Use properties of integrals:
  $$-4\times \int e^{-t} dt$$
- step2: Evaluate the integral:
  $$-4\left(-e^{-t}\right)$$
- step3: Multiply the terms:
  $$4e^{-t}$$
- step4: Add the constant of integral C:
  $$4e^{-t} + C, C \in \mathbb{R}$$
The position equation is $$ s(t) = 4e^{-t} + C $$, where $$ C $$ is a constant of integration.

Given that the initial position is $$ s(0) = 8 $$, we can substitute $$ t = 0 $$ into the position equation to find the value of $$ C $$:
$$ s(0) = 4e^{-0} + C = 4 + C = 8 $$
$$ C = 4 $$

Therefore, the position equation is $$ s(t) = 4e^{-t} + 4 $$.

Now, we have found the position and velocity equations for the object moving along a straight line with the given acceleration, initial velocity, and initial position.
The velocity of the object is $$ v(t) = -4e^{-t} $$ and the position is $$ s(t) = 4e^{-t} + 4 $$.

Find the position and velocity of an object moving along a straight line with the given acceleration, initial velocity,
and initial position.
, and

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