What is the relationship between acceleration and time  a(t)
for the model rocket  (v(t)=αt3+βt+γ
, where  α=−3.0m/s4
,  β=36m/s2
, and  γ=1.0m/s)
?
View Available Hint(s)for Part F

a(t)=α

a(t)=αt2+β+γ

a(t)=3αt2+β

a(t)=c1t2+c2
with the constants  c1
and  c2
unknown because not enough information has been provided
a(t)=β

Previous Answers
Correct
The acceleration of an object is the derivative with respect to time of its velocity-time relationship.
Part G
Using the relationship obtained in Part F, evaluate the acceleration of the model rocket at times  t0=0.0s
,  t1=1.0s
, and  t2=2.0s
.
a(0.0s)
,  a(1.0s)
,  a(2.0s)=

36m/s
,  27m/s
,  0.0m/s

36m/s2
,  27m/s2
,  0.0m/s2

36m/s2
,  36m/s2
,  36m/s2

36m/s2
,  45m/s2
,  72m/s2

### Part F: Finding the Relationship Between Acceleration and Time

Given the velocity function:

\[ v(t) = \alpha t^3 + \beta t + \gamma \]

where:
- \(\alpha = -3.0 \, \text{m/s}^4\)
- \(\beta = 36 \, \text{m/s}^2\)
- \(\gamma = 1.0 \, \text{m/s}\)

The acceleration \(a(t)\) is the derivative of the velocity \(v(t)\) with respect to time \(t\):

\[
a(t) = \frac{d}{dt}v(t)
\]

Let's differentiate \(v(t)\) with respect to \(t\):

\[
v(t) = \alpha t^3 + \beta t + \gamma
\]

\[
a(t) = \frac{d}{dt}(\alpha t^3 + \beta t + \gamma)
\]

\[
a(t) = 3\alpha t^2 + \beta
\]

Thus, the relationship between acceleration and time is:

\[
a(t) = 3\alpha t^2 + \beta
\]

So the correct answer for Part F is:

**\[ a(t) = 3\alpha t^2 + \beta \]**

### Part G: Evaluating the Acceleration at Specific Times

Using the relationship obtained:

\[
a(t) = 3\alpha t^2 + \beta
\]

Substituting the given values of \(\alpha\) and \(\beta\):

\[
a(t) = 3(-3.0) t^2 + 36 = -9t^2 + 36
\]

Now, evaluate the acceleration at the given times:

1. **At \(t_0 = 0.0 \, \text{s}\):**

\[
a(0.0) = -9(0.0)^2 + 36 = 36 \, \text{m/s}^2
\]

2. **At \(t_1 = 1.0 \, \text{s}\):**

\[
a(1.0) = -9(1.0)^2 + 36 = -9(1) + 36 = 27 \, \text{m/s}^2
\]

3. **At \(t_2 = 2.0 \, \text{s}\):**

\[
a(2.0) = -9(2.0)^2 + 36 = -9(4) + 36 = -36 + 36 = 0.0 \, \text{m/s}^2
\]

So the correct answer for Part G is:

**\(a(0.0 \, \text{s}) = 36 \, \text{m/s}^2\), \(a(1.0 \, \text{s}) = 27 \, \text{m/s}^2\), \(a(2.0 \, \text{s}) = 0.0 \, \text{m/s}^2\)**

---

Would you like further details or have any questions?

**Related Questions:**
1. How does the acceleration change if \(\alpha\) is positive?
2. What is the physical significance of the constant \(\gamma\) in the velocity function?
3. How would you find the maximum velocity of the model rocket?
4. How does the velocity change over time for this rocket model?
5. What would the acceleration be at \(t = 3.0 \, \text{s}\)?

**Tip:** The derivative of velocity with respect to time gives acceleration, and similarly, the derivative of acceleration with respect to time gives the jerk, which describes the rate of change of acceleration.

Explore the relationship between acceleration and time for a model rocket using calculus. Learn how to calculate acceleration from a given velocity-time function and evaluate it at specific times. Includes detailed derivations and numerical evaluations.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation