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

2 Answers
Nov 30, 2016

#-4000ms^-2#
#-ve# sign shows that it is retardation or deceleration.

Explanation:

Th kinematic equation connecting quantities of interest is
#v^2-u^2=2as#
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=2a20/100#
Solving for #a#
#a=-40^2xx100/40#
#a=-4000ms^-2#
#-ve# sign shows that it is retardation or deceleration.

Jul 23, 2018

The acceleration is #-4000# #"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+2ad#,

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="0 m/s"#

#v_i="40 m/s"#

#d="20/100 m"#

Unknown

acceleration, #a#

Solution

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

#a=(v_f^2-v_i^2)/(2d)#

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

#a=(-1600"m"^2/"s"^2)/(40/100"m")="# #larr# Invert fraction and multiply.

#a=-1600color(red)cancel(color(black)("m"))^1/"s"^2xx100/(40color(red)cancel(color(black)("m")))=-4000# #"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).