# A car rolling down a hill accelerates from rest to 22m/s in 40s. What is the average velocity of the car? What is the displacement of the car as it rolls down the hill?

Jul 6, 2018

${v}_{\text{average}} = 11 m {s}^{- 1}$

$d = 440 m$

#### Explanation:

The car is rolling down a hill so the acceleration is constant.
This is because acceleration due to gravity is constant, so the acceleration of the car on a given slope will be constant (assuming the slope of the hill is constant and ignoring friction).

equations we know:

1. ${v}_{\text{average}} = \frac{\Delta d}{\Delta t}$
2. ${v}_{\text{average}} = \frac{{v}_{1} - {v}_{0}}{2}$ (where $a$ is constant)
3. ${a}_{\text{average}} = \frac{\Delta v}{\Delta t}$
4. $d = {v}_{\text{average}} \cdot t$
5. $d = {v}_{0} t + \frac{1}{2} a {t}^{2}$ (where $a$ is constant)
6. $\Delta v = \left({v}_{1} - {v}_{0}\right) = 22 - 0 = 22 m {s}^{- 1}$
7. $\Delta t = 40 s$

we know that the car accelerates from $0 \to 22 m {s}^{- 1}$ in 40s so choosing eqn2 and eqn4. :

${v}_{\text{average}} = \frac{22 - 0}{2} = 11 m {s}^{- 1}$

$d = 11 m {s}^{- 1} \cdot 40 s = 440 m$