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3 years ago

how long does it take a car to accelerate from 3.4m/s to 20.9m/s if the average acceleration is 6.0m/s (squared).

Physics

1 answer:

Gre4nikov [31]3 years ago

4 0

Average acceleration is the ratio of the change in velocity to the time it takes for that change to occur:

$a_{\mathrm{av}}=\dfrac{\Delta v}{\Delta t}$

We want to find $\Delta t$ :

$6.0\,\dfrac{\mathrm m}{\mathrm s^2}=\dfrac{20.9\,\frac{\mathrm m}{\mathrm s}-3.4\,\frac{\mathrm m}{\mathrm s}}{\Delta t}$

$\implies\Delta t=2.9\,\mathrm s$

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cestrela7 [59]

Answer:

1. $a_{rad}=17545.2\frac{m}{s^{2}}$

2. $a=4429.45 \frac{m}{s^{2}}$

Explanation:

Radial acceleration is:

$a_{rad}=\frac{v^2}{r}$ (1)

With r the radius respects the axis of rotation and v the tangential velocity that is related with angular velocity (ω) by:

$v= \omega r$ (2)

By (2) on (1):

$a_{rad}= \frac{(\omega r)^2}{r}= (\omega )^2r=(418,88\frac{rad}{s})^2(0.1m)$

$a_{rad}=17545.2\frac{m}{s^{2}}$

To find the acceleration of the tube with the fall, we can use the expression:

$\overrightarrow{J}=\overrightarrow{F}_{avg}(\varDelta t)$ (3)

Due impulse-momentum theorem:

$\overrightarrow{J}=\overrightarrow{p}_{f}-\overrightarrow{p}_{i}$ (4)

with p the momentum and J the impulse. By (4) on (3):

$\overrightarrow{p}_{f}-\overrightarrow{p}_{i}=\overrightarrow{F}_{avg}(\varDelta t)$

And using Newton's second law (F=ma) and that (P=mv):

$mv_f-mv_i=(ma)(\varDelta t)$ (5)

Final velocity is the velocity just after the encounter with hard floor, and initial momentum us just before that moment so the first one is zero and the second one can be found sing conservation of energy:

$\frac{mv_i}{2}=mgh$