We are given the equation of motion:

$S = 2t^3 - 21t^2 + 60t + 12$

where $S$ is the distance from point $O$ and $t$ is the time in seconds.

(a) Distance from O at the start of the journey

At the start of the journey, $t = 0$:

$S(0) = 2(0)^3 - 21(0)^2 + 60(0) + 12 = 12$

Answer: The car is 12 meters from $O$ at the start.

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(b) Times when the car is momentarily at rest

The car is momentarily at rest when its velocity is zero. Velocity $V$ is the first derivative of $S$:

$V = \frac{dS}{dt} = \frac{d}{dt} (2t^3 - 21t^2 + 60t + 12)$

$V = 6t^2 - 42t + 60$

Setting $V = 0$:

$6t^2 - 42t + 60 = 0$

Dividing the equation by 6:

$t^2 - 7t + 10 = 0$

Factoring:

$(t - 5)(t - 2) = 0$

So, $t = 2$ or $t = 5$.

Answer: The car is momentarily at rest at $t = 2$s and $t = 5$s.

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(c) Velocity at the start of the journey

Velocity at $t = 0$:

$V(0) = 6(0)^2 - 42(0) + 60 = 60$

Answer: The initial velocity is 60 m/s.

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(d) Distance covered in the 6th second

The distance covered in the 6th second is:

$S(6) - S(5)$

First, find $S(6)$:

$S(6) = 2(6)^3 - 21(6)^2 + 60(6) + 12$

$= 2(216) - 21(36) + 360 + 12$

$= 432 - 756 + 360 + 12 = 48$

Next, find $S(5)$:

$S(5) = 2(5)^3 - 21(5)^2 + 60(5) + 12$

$= 2(125) - 21(25) + 300 + 12$

$= 250 - 525 + 300 + 12 = 37$

$\text{Distance covered in 6th second} = S(6) - S(5) = 48 - 37 = 11$

Answer: The car covers 11 meters in the 6th second.

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Summary of Answers:

(a) $S = 12$ meters
(b) $t = 2$s and $t = 5$s
(c) $V = 60$ m/s
(d) Distance in 6th second = $11$ meters

Would you like a breakdown of any step?

Here are 5 related questions for practice:

1. Find the acceleration of the car at any time $t$.
2. Determine the total displacement of the car in the first 6 seconds.
3. Find the velocity of the car at $t = 3$s.
4. What is the speed of the car when it returns to its initial position?
5. At what time does the car reach its maximum or minimum distance from $O$?

Tip:

The second derivative of $S$, which is acceleration, tells us if the car is speeding up or slowing down!