# What is the acceleration of a body that goes from "40 m/s" to rest over a distance of "20/100 m"?

Nov 30, 2016

$- 4000 m {s}^{-} 2$
$- v e$ sign shows that it is retardation or deceleration.

#### Explanation:

Th kinematic equation connecting quantities of interest is
${v}^{2} - {u}^{2} = 2 a s$
where $v$ is final velocity, $u$ is initial velocity, $a$ being acceleration and $s$ distance moved.

Inserting given values in the equation in SI units, we get,
${0}^{2} - {40}^{2} = 2 a \frac{20}{100}$
Solving for $a$
$a = - {40}^{2} \times \frac{100}{40}$
$a = - 4000 m {s}^{-} 2$
$- v e$ sign shows that it is retardation or deceleration.

Jul 23, 2018

The acceleration is $- 4000$ $\text{m/s"^2}$.

The acceleration is negative because the object slowed to a stop. This is a negative acceleration, sometimes called deceleration (but not in Physics).

#### Explanation:

Use the equation:

${v}_{f}^{2} = {v}_{i}^{2} + 2 a d$,

where:

${v}_{f}$ is the final velocity, ${v}_{i}$ is the initial velocity, $a$ is the acceleration, and $d$ is the distance.

Known

${v}_{f} = \text{0 m/s}$

${v}_{i} = \text{40 m/s}$

$d = \text{20/100 m}$

Unknown

acceleration, $a$

Solution

Rearrange the equation to isolate $a$. Plug in the known values and solve.

$a = \frac{{v}_{f}^{2} - {v}_{i}^{2}}{2 d}$

a=((0)-(40"m"/"s")^2)/(2*20/100"m")="

a=(-1600"m"^2/"s"^2)/(40/100"m")=" $\leftarrow$ Invert fraction and multiply.

a=-1600color(red)cancel(color(black)("m"))^1/"s"^2xx100/(40color(red)cancel(color(black)("m")))=-4000 $\text{m/s"^2}$

The acceleration is negative because the object slowed to a stop. This is a negative acceleration, sometimes called deceleration (but not in Physics).