# Question #c57c0

Mar 5, 2016

Yes.

#### Explanation:

Force $F = m \cdot a$
Where $m$ is the mass and $a$ its acceleration of the object.
Also Kinetic Energy $K E = \frac{1}{2} m {v}^{2}$
Given that $F \propto \frac{1}{v}$
We know that acceleration $a$ is $\equiv \frac{\mathrm{dv}}{\mathrm{dt}}$
$\therefore$ one of the $v$ in the expression for Kinetic energy can be written as
$K E = \frac{1}{2} m {v}^{2} \propto m \left(a . t\right) . v$
Since the factor $\frac{1}{2}$ is a constant, it has been absorbed in the propotionality. Rearranging

$K E \propto \left(m . a\right) . \left(t . v\right)$

Substituting for force and its given inverse proportionality we get
$K E \propto \frac{1}{\cancel{v}} . \left(t . \cancel{v}\right)$
$K E \propto t$

$t$ is time

Does this help?
{"ft_lang_label":"__label__en","ft_lang_prob":0.6202243,"math_prob":0.9999645,"size":359,"snap":"2019-35-2019-39","text_gpt3_token_len":88,"char_repetition_ratio":0.10422535,"word_repetition_ratio":0.0,"special_character_ratio":0.20612814,"punctuation_ratio":0.08064516,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000031,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-20T12:21:19Z\",\"WARC-Record-ID\":\"<urn:uuid:2eb59619-dc95-4597-8a28-1e1564ba66b8>\",\"Content-Length\":\"33374\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:17f792f5-1973-4582-83ae-a17869ce017a>\",\"WARC-Concurrent-To\":\"<urn:uuid:bed0b100-36b3-4b93-ab30-2dfc37f5102f>\",\"WARC-IP-Address\":\"54.221.217.175\",\"WARC-Target-URI\":\"https://socratic.org/questions/56da9c317c014979b68c57c0\",\"WARC-Payload-Digest\":\"sha1:XI2TCPRPD2UXEQXIDGRZGSS47A5A2HP2\",\"WARC-Block-Digest\":\"sha1:6TVLHPKXNKQ45Q7UVWXHPY3WBJ6CHVX4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514574018.53_warc_CC-MAIN-20190920113425-20190920135425-00379.warc.gz\"}"}