Question 1 A car moved distance S meters in a time t seconds, according to the formula \( S=3 t^{3}-2 t^{2}+4 t-1 \). Determine: 1.1 velocity at (i) \( t=0,(i i) t=1.5 \mathrm{~s} \). 1.2 Acceleration at (i) \( t=0,(i i) t=1.5 \mathrm{~s} \).

Alright, I need to determine the velocity and acceleration of a car at specific times given the distance formula \( S = 3t^3 - 2t^2 + 4t - 1 \). Let's break this down step by step. First, I recall that velocity is the first derivative of distance with respect to time, and acceleration is the second derivative of distance with respect to time. So, I'll need to find the first and second derivatives of the given function. Starting with the distance function: \[ S(t) = 3t^3 - 2t^2 + 4t - 1 \] **Finding Velocity:** To find the velocity \( v(t) \), I'll take the first derivative of \( S(t) \) with respect to time \( t \): \[ v(t) = \frac{dS}{dt} = \frac{d}{dt}(3t^3) - \frac{d}{dt}(2t^2) + \frac{d}{dt}(4t) - \frac{d}{dt}(1) \] \[ v(t) = 9t^2 - 4t + 4 \] **Finding Acceleration:** Next, to find the acceleration \( a(t) \), I'll take the second derivative of \( S(t) \) with respect to time \( t \): \[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(9t^2) - \frac{d}{dt}(4t) + \frac{d}{dt}(4) \] \[ a(t) = 18t - 4 \] Now, I need to evaluate these derivatives at the given times: \( t = 0 \) and \( t = 1.5 \) seconds. **At \( t = 0 \) seconds:** - **Velocity:** \[ v(0) = 9(0)^2 - 4(0) + 4 = 0 - 0 + 4 = 4 \, \text{m/s} \] - **Acceleration:** \[ a(0) = 18(0) - 4 = 0 - 4 = -4 \, \text{m/s}^2 \] **At \( t = 1.5 \) seconds:** - **Velocity:** \[ v(1.5) = 9(1.5)^2 - 4(1.5) + 4 \] First, calculate \( (1.5)^2 = 2.25 \) \[ v(1.5) = 9(2.25) - 4(1.5) + 4 = 20.25 - 6 + 4 = 18.25 \, \text{m/s} \] - **Acceleration:** \[ a(1.5) = 18(1.5) - 4 = 27 - 4 = 23 \, \text{m/s}^2 \] Wait a second, acceleration at \( t = 1.5 \) seconds is 23 m/s², which seems quite high. Is that physically plausible? Let me double-check the calculations. Recalculating \( v(1.5) \): \[ v(1.5) = 9(2.25) - 4(1.5) + 4 = 20.25 - 6 + 4 = 18.25 \, \text{m/s} \] That seems correct. And \( a(1.5) = 23 \, \text{m/s}^2 \) is indeed a high acceleration, but mathematically, it's correct based on the given function. Perhaps the context of the problem allows for such high acceleration, or it's a theoretical scenario. **Summary:** - At \( t = 0 \) seconds: - Velocity: 4 m/s - Acceleration: -4 m/s² - At \( t = 1.5 \) seconds: - Velocity: 18.25 m/s - Acceleration: 23 m/s² I should also consider the units to ensure consistency. The distance \( S \) is in meters, and time \( t \) is in seconds, so velocity \( v \) will be in meters per second (m/s), and acceleration \( a \) will be in meters per second squared (m/s²), which matches the units I've calculated. Additionally, I might want to verify if there are any constraints or real-world limitations that could affect these values, but based on the given formula, these are the accurate results.