piarepm

2021-12-17

When jumping, a flea reaches a takeoff speed of 1.0 m/s over a distance of 0.50 mm.
a. What is the fleas

SlabydouluS62

Step 1
Given values:
${v}_{iy}=0\frac{m}{s}$
${v}_{fy}=1\frac{m}{s}$
${y}_{i}=0m$
${y}_{f}=0.5\cdot {10}^{-3}m$
$g=9.81\frac{m}{{s}^{2}}$
a) In this case, we have to determine value of acceleration. The best way to find that is using the kinematic equation:
${v}_{f}^{2}={v}_{i}^{2}+2a\left(\mathrm{△}y\right)$.
Plug in values in above equation and solve for acceleratiion:
${v}_{fy}^{2}={v}_{iy}^{2}+2{a}_{y}\left({y}_{f}-{y}_{i}\right)$
${\left(1\frac{m}{s}\right)}^{2}=\left(0\frac{m}{s}\right)+2{a}_{y}\left(0.5\cdot {10}^{-3}m-0m\right)$
$\left(1\frac{{m}^{2}}{{s}^{2}}=2{a}_{y}\left(0.5\cdot {10}^{-3}m\right)$
${a}_{y}=\frac{\left(1\frac{{m}^{2}}{{s}^{2}}\right)}{2\left(0.5\cdot {10}^{-3}m\right)}$
${a}_{y}=1000\frac{m}{{s}^{2}}$
Step 2
b) Use the kinematic equation to find value of time:
${v}_{fy}={v}_{iy}+{a}_{y}t$
Plug in values in above equation and solve for time:
$1\frac{m}{s}=0\frac{m}{s}+\left(1000\frac{m}{{s}^{2}}\right)t$
$t=\frac{1\frac{m}{s}-0\frac{m}{s}}{\left(1000\frac{m}{{s}^{2}}\right)}$
$t=\frac{1\frac{m}{s}}{\left(1000\frac{m}{{s}^{2}}\right)}$
$t={10}^{-3}s$
c) It is moving straight up, so it's final speed is equal to zero when it reaches the highest point is using the kinematic equation:
${v}_{f}^{2}={v}_{i}^{2}+2a\left(\mathrm{△}y\right)$
If we plug in values in above equation, we get:
${v}_{fy}^{2}={v}_{iy}^{2}+2{a}_{y}\left({y}_{f}-{y}_{i}\right)$

Donald Cheek

Step 1
Part A.
Following equation will be used:
${v}^{2}-{u}^{2}=2as$
where v is the final velocity , u is the initial velocity, a is the acceleration and s is the distance traveled
Given values are as follows:
$v=1\frac{m}{s},u=0\frac{m}{s},s=0.5×{10}^{-3}m$
Therefore, ${v}^{2}-{u}^{2}=2as$
${\left(1\right)}^{2}-{\left(0\right)}^{2}=2×a×5×{10}^{-3}$
$a=100\frac{m}{{s}^{2}}$
Step 2
Part B.
Here time is need to be calculated by following equation:
$v=u+at$
$1=0+100×$
$t=0.01s$
Part C.
If the flea jumps straight up then the acceleration due to gravity will be taken into account and following equation will be used.
${v}^{2}-{u}^{2}=2gs$
Given values are as follows:
$v=0\frac{m}{s},u=1\frac{m}{s},g=-9.8\frac{m}{{s}^{2}}$
Therefore, ${v}^{2}-{u}^{2}=2gs$
${\left(0\right)}^{2}-{\left(1\right)}^{2}=-2×9.8×s$
$s=0.051m$

Nick Camelot

We have:
$a=\frac{{v}^{2}-{u}^{2}}{2s}$
where:
$a$ is the average acceleration,
$v$ is the final velocity,
$u$ is the initial velocity, and
$s$ is the displacement.
Given that the initial velocity $u$ is 0 m/s (since the flea starts from rest), the final velocity $v$ is 1.0 m/s, and the displacement $s$ is 0.50 mm (or 0.50 × ${10}^{-3}$ m), we can substitute these values into the equation:
$a=\frac{\left(1.0\phantom{\rule{0.167em}{0ex}}\text{m/s}{\right)}^{2}-\left(0\phantom{\rule{0.167em}{0ex}}\text{m/s}{\right)}^{2}}{2×\left(0.50×{10}^{-3}\phantom{\rule{0.167em}{0ex}}\text{m}\right)}$
Simplifying further:
$a=\frac{1.0\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}}{1.0×{10}^{-3}\phantom{\rule{0.167em}{0ex}}\text{m}}$
$a=1.0×{10}^{3}\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$
Therefore, the flea's average acceleration is $1.0×{10}^{3}\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$.

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