$s = 2t^4 - \frac{1}{6}t^3 + 2t^2$

$v = \dfrac{ds}{dt}$

$v = 8t^3 - \frac{1}{2}t^2 + 4t$

$a = \dfrac{dv}{dt}$

$a = 24t^2 - t + 4$

When t = 2 sec

$v = 8(2^3) - \frac{1}{2}(2^2) + 4(2)$
$v = 70 \, \text{ m/s}$ *answer*

$a = 24(2^2) - 2 + 4$

$a = 98\, \text{ m/s}^2$ *answer*

## Comments

## A train travel between two

A train travel between two stations 1/2 miles apart in minimum time of 41 seconds if the train accelerate and decelerate 8 feet per second squared starting at rest at the first station and coming to a stop at second station what is the minimum speed in MPH?