Math Problem Statement
Suppose that the position of a particle is given by s(t) = 3t^3 + 5t + 9. (a) Find the velocity at time t. (b) Find the velocity at time t = 3 seconds. (c) Find the acceleration at time t. (d) Find the acceleration at time t = 3 seconds.
Solution
Let's analyze the problem step by step.
The position of the particle is given by .
(a) Find the velocity at time .
The velocity is the first derivative of the position function with respect to time : Given , let's differentiate:
= 9t^2 + 5$$ So, the velocity function is: $$v(t) = 9t^2 + 5 \, \frac{m}{s}$$ ### (b) Find the velocity at $$t = 3$$ seconds. We substitute $$t = 3$$ into the velocity function: $$v(3) = 9(3)^2 + 5 = 9(9) + 5 = 81 + 5 = 86 \, \frac{m}{s}$$ ### (c) Find the acceleration at time $$t$$. The acceleration is the derivative of the velocity function $$v(t)$$ with respect to time $$t$$: $$a(t) = \frac{dv(t)}{dt}$$ Since $$v(t) = 9t^2 + 5$$, let's differentiate: $$a(t) = \frac{d}{dt} \left( 9t^2 + 5 \right) = 18t$$ So, the acceleration function is: $$a(t) = 18t \, \frac{m}{s^2}$$ ### (d) Find the acceleration at $$t = 3$$ seconds. We substitute $$t = 3$$ into the acceleration function: $$a(3) = 18(3) = 54 \, \frac{m}{s^2}$$ ### Summary of Answers: - (a) $$v(t) = 9t^2 + 5 \, \frac{m}{s}$$ - (b) $$v(3) = 86 \, \frac{m}{s}$$ - (c) $$a(t) = 18t \, \frac{m}{s^2}$$ - (d) $$a(3) = 54 \, \frac{m}{s^2}$$ Would you like any further explanation or details on these steps? --- Here are 5 related questions to practice: 1. What would the velocity be at $$t = 5$$ seconds? 2. How would the position function look if the velocity was $$v(t) = 6t^2 + 4t$$? 3. How do you find the jerk (third derivative of position) in this case? 4. What is the significance of the constant 9 in the position equation? 5. How would you determine when the particle is at rest? **Tip**: Always remember to differentiate the position function to find velocity and differentiate the velocity function to find acceleration!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Calculus
Derivatives
Velocity
Acceleration
Formulas
v(t) = ds(t)/dt
a(t) = dv(t)/dt
Theorems
Derivative rules in calculus
Suitable Grade Level
Grades 11-12
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