3 Speedy Gonzales runs from point A in a straight line, coming to rest at point B. The distance in a straight…

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3 Speedy Gonzales runs from point A in a straight line, coming to rest at point B. The distance in a straight line from A is given by the formula s = 3t3 - 1/2t4 where t is measure

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2.3 Speedy Gonzales runs from point A in a straight line, coming to rest at point B. The distance in a straight line from A is given by the formula $s = 3t^3 - \frac{1}{2}t^4$ where $t$ is measured in seconds and $s$ in metres.

2.3.1 An expression for the velocity of Speedy Gonzales in terms of $t$
Velocity $v$ is the derivative of distance $s$ with respect to time $t$ .

Differentiate the given distance function $s$ with respect to $t$ .
$v = \frac{ds}{dt} = \frac{d}{dt}\left(3t^3 - \frac{1}{2}t^4\right)$ $v = 3(3t^{3-1}) - \frac{1}{2}(4t^{4-1})$ $v = 9t^2 - 2t^3$
The expression for velocity is $\boxed{v = 9t^2 - 2t^3}$ .

2.3.2 The time taken to reach B
Speedy Gonzales comes to rest at point B, which means its velocity $v$ is 0.
Set the velocity expression from 2.3.1 to zero.
$9t^2 - 2t^3 = 0$
Factor out the common term $t^2$ .
$t^2(9 - 2t) = 0$
Solve for $t$ .
This gives two possible solutions: $t^2 = 0$ or $9 - 2t = 0$ .
$t=0$ represents the starting point A.
For $9 - 2t = 0$ :
$2t = 9$ $t = \frac{9}{2} = 4.5 \text{ seconds}$
The time taken to reach B is $\boxed{4.5 \text{ seconds}}$ .

**2.

3 Speedy Gonzales runs from point A in a straight line, coming to rest at point B. The distance in a straight line from A is given by the formula s = 3t3 − (1)/(2)t4 where t is measure

Step-by-step solution

Differentiate the given distance function $s$ with respect to $t$ .

Set the velocity expression from 2.3.1 to zero.

Factor out the common term $t^2$ .

Solve for $t$ .

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