Question

A van slows down uniformly from 17 m/s to 0 m/s in 5 s. How far does it travel before stopping?

Procedure is required.

Answer 1

Answer:

Before stopping, the van travels 42.5 m

Explanation:

Constant Acceleration Motion

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

$v_f=v_o+at\qquad\qquad [1]$

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]$

The van slows down uniformly from v0=17 m/s to vf=0 m/s in t=5 s. The acceleration can be calculated by solving [1] for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

$\displaystyle a=\frac{0-17}{5}=-3.4\ m/s^2$

The distance covered is:

$\displaystyle x=17\cdot 5+\frac{(-3.4)\cdot 5^2}{2}$

$\displaystyle x=85-42.5=42.5$

Before stopping, the van travels 42.5 m