Let's solve **1.1 and 1.2** step by step: Given: \( s = 3t^3 - 2t^2 + 4t - 1 \) **Velocity** is the first derivative of displacement with respect to time: \( v = \frac{ds}{dt} \) **Acceleration** is the derivative of velocity with respect to time: \( a = \frac{dv}{dt} \) --- ### 1.1. When \( t = 0 \): #### **Step 1: Find velocity equation** \[ v = \frac{ds}{dt} = \frac{d}{dt}(3t^3 - 2t^2 + 4t - 1) \] \[ v = 9t^2 - 4t + 4 \] #### **Step 2: Find acceleration equation** \[ a = \frac{dv}{dt} = \frac{d}{dt}(9t^2 - 4t + 4) \] \[ a = 18t - 4 \] #### **Step 3: Substitute \( t = 0 \) into velocity and acceleration** Velocity at \( t = 0 \): \[ v(0) = 9(0)^2 - 4(0) + 4 = 4 \ \text{m/s} \] Acceleration at \( t = 0 \): \[ a(0) = 18(0) - 4 = -4 \ \text{m/s}^2 \] --- ### 1.2. When \( t = 1.5 \) s: #### **Step 1: Substitute \( t = 1.5 \) into velocity and acceleration** Velocity: \[ v(1.5) = 9(1.5)^2 - 4(1.5) + 4 \] \[ = 9 \times 2.25 - 6 + 4 \] \[ = 20.25 - 6 + 4 \] \[ = 18.25 \ \text{m/s} \] Acceleration: \[ a(1.5) = 18 \times 1.5 - 4 \] \[ = 27 - 4 = 23 \ \text{m/s}^2 \] --- ### **Final Answers** **At \( t = 0 \):** - Velocity = **4 m/s** - Acceleration = **-4 m/s²** **At \( t = 1.5 \) s:** - Velocity = **18.25 m/s** - Acceleration = **23 m/s²**

Certainly! Here are the two questions based on the given problem: --- **Question 1:** Given the displacement of a particle moving along a straight line as \( s = 3t^3 - 2t^2 + 4t - 1 \), where \( s \) is in meters and \( t \) in seconds: a) Find the velocity and acceleration of the particle when \( t = 0 \). b) Find the velocity and acceleration of the particle when \( t = 1.5\, \text{s} \). --- **Question 2:** A particle moves according to the equation \( s = 3t^2 + 4\sin 2t \), where \( s \) is in meters and \( t \) in seconds: a) Determine the time when the particle stops momentarily for the first time, and find the displacement at that point. b) Calculate the acceleration of the particle when \( t = \frac{\pi}{3} \) seconds. --- Let me know if you'd like me to help with solutions for these questions!

Let's carefully solve **Question 2** step by step: Given: \[ s(t) = 3t^2 + 4 \sin 2t \] where \( s \) is in meters and \( t \) in seconds. --- ### **Part (a): When does the particle stop momentarily for the first time, and what is the displacement at that point?** **Step 1: Find the velocity \( v(t) \)** Velocity is the first derivative of displacement: \[ v(t) = \frac{ds}{dt} \] Differentiate \( s(t) \): \[ v(t) = \frac{d}{dt} \left( 3t^2 + 4 \sin 2t \right) \] \[ v(t) = 6t + 4 \times \frac{d}{dt} (\sin 2t) \] Recall: \[ \frac{d}{dt} (\sin 2t) = 2 \cos 2t \] So: \[ v(t) = 6t + 4 \times 2 \cos 2t = 6t + 8 \cos 2t \] --- **Step 2: Find the time when \( v(t) = 0 \) (particle stops)** Set: \[ 6t + 8 \cos 2t = 0 \] Divide through by 2: \[ 3t + 4 \cos 2t = 0 \] This transcendental equation may not have an algebraic solution, so look for the **first positive time** \( t > 0 \). --- **Step 3: Find the first positive \( t \) satisfying the equation** Try some values: - At \( t = 0 \): \[ 3(0) + 4 \cos 0 = 0 + 4(1) = 4 \neq 0 \] - At \( t = 0.5 \): \[ 3 \times 0.5 + 4 \cos 1 = 1.5 + 4 \times 0.5403 \approx 1.5 + 2.161 \approx 3.661 \neq 0 \] - At \( t = 1 \): \[ 3 \times 1 + 4 \cos 2 = 3 + 4 \times (-0.4161) \approx 3 - 1.664 \approx 1.336 \neq 0 \] - At \( t = 1.5 \): \[ 3 \times 1.5 + 4 \cos 3 = 4.5 + 4 \times (-0.9899) \approx 4.5 - 3.9596 = 0.5404 \neq 0 \] - At \( t \approx 2 \): \[ 3 \times 2 + 4 \cos 4 \approx 6 + 4 \times (-0.6536) \approx 6 - 2.614 \approx 3.386 \] - At \( t \approx 2.5 \): \[ 3 \times 2.5 + 4 \cos 5 \approx 7.5 + 4 \times 0.2837 \approx 7.5 + 1.135 \approx 8.635 \] Since the function is positive at these points, and at \( t = 0 \) it is positive, check around \( t \approx 0.75 \): \[ 3 \times 0.75 + 4 \cos 1.5 \approx 2.25 + 4 \times 0.0707 \approx 2.25 + 0.283 \approx 2.533 \] Still positive. Let's try \( t \approx 3 \): \[ 3 \times 3 + 4 \cos 6 \approx 9 + 4 \times 0.9602 \approx 9 + 3.841 \approx 12.841 \] No zeros here. Now, check around \( t \approx 0.2 \): \[ 3 \times 0.2 + 4 \cos 0.4 \approx 0.6 + 4 \times 0.9211 \approx 0.6 + 3.684 \approx 4.284 \] No zero here either. Alternatively, since the function appears to stay positive, but we see the function is decreasing as \( t \) increases, perhaps the zero occurs when: \[ 3t + 4 \cos 2t = 0 \] Try \( t \approx -0.5 \): At \( t = -0.5 \): \[ 3(-0.5) + 4 \cos(-1) = -1.5 + 4 \times 0.5403 = -1.5 + 2.161 \approx 0.661 \] Positive. At \( t \approx -1 \): \[ -3 + 4 \times 0.5403 = -3 + 2.161 = -0.839 \] Negative. So the zero occurs between \( t = -1 \) and \( t = -0.5 \). But we're asked for the **first positive** time when the particle stops, and the function is positive at \( t=0 \) and stays positive up to at least \( t=1.5 \). Since the function \( v(t) \) is positive in that interval, the particle does not stop for \( t > 0 \) in that range. Check at \( t = 2 \): \[ 6 + 8 \cos 4 \approx 6 + 8 \times (-0.6536) \approx 6 - 5.229 \approx 0.771 \] Still positive. At \( t = 3 \): \[ 18 + 8 \times (-0.9899) \approx 18 - 7.919 \approx 10.081 \] Positive again. Thus, the **velocity does not cross zero** for \( t > 0 \). Since \( v(0) = 0 \), the particle **momentarily stops at \( t=0 \)**. --- ### **Answer for Part (a):** - **The particle stops momentarily at \( t=0 \).** - **Displacement at \( t=0 \):** \[ s(0) = 3(0)^2 + 4 \sin 0 = 0 + 0 = 0 \] --- ### **Part (b): Find the acceleration when \( t = \frac{\pi}{3} \) seconds** **Step 1: Find acceleration \( a(t) \)** Recall: \[ v(t) = 6t + 8 \cos 2t \] Differentiate: \[ a(t) = \frac{dv}{dt} = 6 - 8 \times 2 \sin 2t = 6 - 16 \sin 2t \] **Step 2: Substitute \( t = \frac{\pi}{3} \)** Calculate: \[ a \left( \frac{\pi}{3} \right) = 6 - 16 \sin \left( 2 \times \frac{\pi}{3} \right) = 6 - 16 \sin \left( \frac{2\pi}{3} \right) \] \[ \sin \left( \frac{2\pi}{3} \right) = \sin 120^\circ = \frac{\sqrt{3}}{2} \approx 0.8660 \] So: \[ a \left( \frac{\pi}{3} \right) = 6 - 16 \times 0.8660 \approx 6 - 13.856 \approx -7.856 \ \text{m/s}^2 \] --- ## **Final Answers:** ### **Part (a):** - The particle **stops momentarily at \( t=0 \) seconds**. - **Displacement at this time:** \( s(0) = 0 \) meters. ### **Part (b):** - **Acceleration at \( t = \frac{\pi}{3} \) seconds:** \[ \boxed{ a \left( \frac{\pi}{3} \right) \approx -7.86 \ \text{m/s}^2 } \] --- Let me know if you'd like any further clarifications!