When a cyclist squeezes her brake lever, she can stop wlth an acceleration of $3.0$ $m$ $/$ $s^2$ . How far will her bike travel while coming to a complete step if her initial speed was $11$ $m$ $/$ $s$ ?

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When a cyclist squeezes her brake lever, she can stop wlth an acceleration of $3.0$ $m$ $/$ $s^2$ . How far will her bike travel while coming to a complete step if her initial speed was $11$ $m$ $/$ $s$ ?

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Answer 1 · 2015-12-03T03:27:04

I found: $20.2m$

Explanation:

Here you can use the relationship from kinematics:
$v_f^2=v_i^2+2ad$
Where $f and i$ refer to the initial and final positions:
with your data and taking "d" as the distance up to $v_f=0$ you get:
$0=11^2-2(3)d$
(negative acceleration)
$d=121/6=20.2m$

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When a cyclist squeezes her brake lever, she can stop wlth an acceleration of $3.0$ $m$ $/$ $s^2$ . How far will her bike travel while coming to a complete step if her initial speed was $11$ $m$ $/$ $s$ ?

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Explanation:

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When a cyclist squeezes her brake lever, she can stop wlth an acceleration of 3.03.0 mm//s^2s^2 . How far will her bike travel while coming to a complete step if her initial speed was 1111 mm//ss?

1 Answer

Explanation:

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When a cyclist squeezes her brake lever, she can stop wlth an acceleration of $3.0$ $m$ $/$ $s^2$ . How far will her bike travel while coming to a complete step if her initial speed was $11$ $m$ $/$ $s$ ?