# The position of an object moving along a line is given by p(t) = 2t^3 - 2t +2. What is the speed of the object at t = 4 ?

Dec 18, 2017

#### Answer:

$94 m {s}^{- 1}$

#### Explanation:

$p \left(t\right) = 2 {t}^{3} - 2 t + 2$

to find the speed we differentiate

$p ' \left(t\right) = 6 {t}^{2} - 2$

for $t = 2$

$p ' \left(4\right) = 6 \times {4}^{2} - 2$

speed$= 94 m {s}^{- 1}$

SI units assumed

Dec 18, 2017

#### Answer:

The speed is $= 94 m {s}^{-} 1$

#### Explanation:

The speed of an object is the derivative of the position.

$v \left(t\right) = \frac{\mathrm{dp}}{\mathrm{dt}}$

The position is

$p \left(t\right) = 2 {t}^{3} - 2 t + 2$

The speed is

$v \left(t\right) = p ' \left(t\right) = 6 {t}^{2} - 2$

And when $t = 4$

$v \left(4\right) = 6 \cdot {\left(4\right)}^{2} - 2 = 96 - 2 = 94 m {s}^{-} 1$